download chapter 4: additional derivative topics

216

I N T R O D U C T I O N

In this chapter, we complete our discussion of derivatives by first looking at the dif-

ferentiation of exponential forms. Next, we will consider the differentiation of loga-

rithmic forms. Finally, we will examine some additional topics and applications

involving all the different types of functions we have encountered thus far, including

the general chain rule. You will probably find it helpful to review some of the

important properties of the exponential and logarithmic functions given in Chapter 2

before proceeding further.

4-1 The Constant e andContinuous CompoundInterest

4-2 Derivatives of Exponentialand Logarithmic Functions

4-3 Derivatives of Products and Quotients

4-4 The Chain Rule

4-5 Implicit Differentiation

4-6 Related Rates

4-7 Elasticity of Demand

Chapter 4 Review

Review Exercise

Additional Derivative Topics

4CHAPTER

BARNMC04_0132328186.QXD 2/21/07 1:27 PM Page 216

S e c t i o n 4 - 1 The Constant e and Continuous Compound Interest 217

Section 4-1 THE CONSTANT e AND CONTINUOUSCOMPOUND INTEREST� The Constant e� Continuous Compound Interest

In Chapter 2, both the exponential function with base e and continuous compoundinterest were introduced informally. Now, with limit concepts at our disposal, we cangive precise definitions of e and continuous compound interest.

� The Constant eThe special irrational number e is a particularly suitable base for both exponential andlogarithmic functions. The reasons for choosing this number as a base will becomeclear as we develop differentiation formulas for the exponential function and thenatural logarithmic function ln x.

In precalculus treatments (Chapter 2), the number e is informally defined as theirrational number that can be approximated by the expression by takingn sufficiently large. Now we will use the limit concept to formally define e as eitherof the following two limits:

31 + 11>n24n

ex

f (s) � (1 � s)1/s

s

f (s)

0.50�0.5

1

2

3

4

FIGURE 1

s 0.1 0.2 0.5

4.0000 3.0518 2.8680 2.5937 2.4883 2.25002.7320 : e ; 2.704811 + s21>s-0.01 : 0 ; 0.01-0.1-0.2-0.5

We will use both of these limit forms. [Note: If then, as ]The proof that the indicated limits exist and represent an irrational number

between 2 and 3 is not easy and is omitted here.

n : q , s : 0.s = 1>n,

DEFINITION The Number e

e = 2.718 281 828 459 Á

e = limn: q

a1 +

1nbn

or, alternatively, e = lims:011 + s21>s

The two limits used to define e are unlike any we have encountered thus far. Some peo-ple reason (incorrectly) that both limits are 1, since as and 1 to any poweris 1. A calculator experiment on an ordinary scientific calculator with a key can con-vince you otherwise. Consider the following table of values for s and andthe graph shown in Figure 1 for s close to 0.

f1s2 = 11 + s21>syx

s : 01 + s : 1

I N S I G H T

s approaches O from the left s approaches O from the right: O ;

BARNMC04_0132328186.QXD 2/21/07 1:27 PM 17

Compute some of the table values with a calculator yourself, and also try severalvalues of s even closer to 0. Note that the function is discontinuous at

Exactly who discovered e is still debated. It is named after the great mathemati-cian Leonhard Euler (1707–1783), who computed e to 23 decimal places, using

� Continuous Compound InterestNow we will see how e appears quite naturally in the important application of com-pound interest. Let us start with simple interest, move on to compound interest, andthen proceed on to continuous compound interest.

On the one hand, if a principal P is borrowed at an annual rate r,* then after tyears at simple interest, the borrower will owe the lender an amount A given by

Simple interest (1)

On the other hand, if interest is compounded n times a year, the borrower will owethe lender an amount A given by

Compound interest (2)

where r�n is the interest rate per compounding period and nt is the number of com-pounding periods. Suppose that P, r, and t in equation (2) are held fixed and n isincreased. Will the amount A increase without bound, or will it tend to approachsome limiting value?

Let us perform a calculator experiment before we attack the general limit prob-lem. If and then

We compute A for several values of n in Table 1. The biggest gain appears in the firststep; then the gains slow down as n increases. In fact, it appears that A tends toapproach $112.75 as n gets larger and larger.

A = 100a1 +

0.06nb2n

t = 2 years,P = $100, r = 0.06,

A = Pa1 +

rnbnt

A = P + Prt = P11 + rt2

31 + 11>n24n.

s = 0.

218 C H A P T E R 4 Additional Derivative Topics

* If r is the interest rate written as a decimal, then 100r% is the rate in percent. For example, if we have The expressions 0.12 and 12% are therefore equivalent. Unlessstated otherwise, all formulas in this book use r in decimal form.

100r% = 10010.122% = 12%.r = 0.12,

Compounding Frequency n

Annually 1 $112.3600

Semiannually 2 112.5509

Quarterly 4 112.6493

Monthly 12 112.7160

Weekly 52 112.7419

Daily 365 112.7486

Hourly 8,760 112.7496

A = 100a1 �0.06

nb2n

TABLE 1

Explore & Discuss 1 (A) Suppose that $1,000 is deposited in a savings account that earns 6% sim-ple interest. How much will be in the account after 2 years?

(B) Suppose that $1,000 is deposited in a savings account that earns compoundinterest at a rate of 6% per year. How much will be in the account after 2 yearsif interest is compounded annually? Semiannually? Quarterly? Weekly?

(C) How frequently must interest be compounded at the 6% rate in order tohave $1,150 in the account after 2 years?



Now we turn back to the general problem for a moment. Keeping P, r, and t fixedin equation (2), we compute the following limit and observe an interesting and usefulresult:

= Pert

= P3lims:011 + s21>s4rt

= P lims:0311 + s21>s4rt

limn: q

Pa1 +

rnbnt

= P limn: q

a1 +

rnb1n>r2rt

* The following new limit property is used: If exists, thenprovided that the last expression names a real number.† Following common usage, we will often write “at 6% compounded continuously,” understanding thatthis means “at an annual nominal rate of 6% compounded continuously.”

limx:c3f1x24p = 3limx:c f1x24p,limx:c f1x2

Insert r/r in the exponent andlet Note that

Use a limit property.*

lims:011 + s21>s = e

n : q implies s : 0.s = r>n.

The resulting formula is called the continuous compound interest formula, a veryimportant and widely used formula in business and economics.

Continuous Compound Interest

where

A = amount at time t

t = time in years

r = annual nominal interest rate compounded continuously

P = principal

A = Pert

E X A M P L E 1 Computing Continuously Compounded Interest If $100 is invested at 6%compounded continuously,† what amount will be in the account after 2 years? Howmuch interest will be earned?

SOLUTION

6% is equivalent to

(Compare this result with the values calculated in Table 1.) The interest earned is$112.7497 - $100 = $12.7497.

L $112.7497

r = 0.06. = 100e10.062122 A = Pert

MATCHED PROBLEM 1 What amount (to the nearest cent) will an account have after 5 years if $100 isinvested at an annual nominal rate of 8% compounded annually? Semiannually?Continuously?

E X A M P L E 2 Graphing the Growth of an Investment Recently, Union Savings Bank offereda 5-year certificate of deposit (CD) that earned 5.75% compounded continuously. If$1,000 is invested in one of these CDs, graph the amount in the account as a functionof time for a period of 5 years.

SOLUTION We want to graph

Using a calculator, we construct a table of values (Table 2). Then we graph the pointsfrom the table and join the points with a smooth curve (Fig. 2).

A = 1,000e0.0575t 0 … t … 5



t A($)

0 1,000

1 1,059

2 1,122

3 1,188

4 1,259

5 1,333

TABLE 2

5

500

0

1000

1500

A

t

FIGURE 2

TABLE 3

Depending on the domain, the graph of an exponential function can appear to be linear.Table 3 shows the slopes of the line segments connecting the dots in Figure 2. Since theseslopes are not identical, this graph is not the graph of a straight line.

I N S I G H T

MATCHED PROBLEM 2 If $5,000 is invested in a Union Savings Bank 4-year CD that earns 5.61%compounded continuously, graph the amount in the account as a function of time fora period of 4 years.

E X A M P L E 3 Computing Growth Time How long will it take an investment of $5,000 to growto $8,000 if it is invested at 5% compounded continuously?

SOLUTION Starting with the continous compound interest formula we must solvefor t:

Divide both sides by 5,000 and reverse the equation.

t L 9.4 years

t =

ln 1.60.05

0.05t = ln 1.6

ln e0.05t= ln 1.6

e0.05t= 1.6

8,000 = 5,000e0.05t

A = Pert

A = Pert,

Take the natural logarithm of both sides—recall thatlogb b x

= x.

FIGURE 3

y2 = 8,000 y1 = 5,000e0.05x Figure 3 shows an alternative method for solving Example 3 on a graphing

calculator.

t A m

0 1,000

1 1,059

2 1,122

3 1,188

4 1,259

5 1,333F 74

F 71

F 66

F 63

F 59



Explore & Discuss 2 You are considering three options for investing $10,000: at 7% compoundedannually, at 6% compounded monthly, and at 5% compounded continuously.

(A) Which option would be the best for investing $10,000 for 8 years?

(B) How long would you need to invest your money for the third option to bethe best?

Answers to Matched Problems 1. $146.93; $148.02; $149.18

2. 3. 4.51 yr 4. 19.97 yrA = 5,000e0.0561t

t A($)

0 5,000

1 5,289

2 5,594

3 5,916

4 6,258

40

6000

5000

4000

3000

2000

1000

7000

t

A

MATCHED PROBLEM 3 How long will it take an investment of $10,000 to grow to $15,000 if it is invested at9% compounded continuously?

E X A M P L E 4 Computing Doubling Time How long will it take money to double if it is investedat 6.5% compounded continuously?

SOLUTION Starting with the continuous compound interest formula we must solvefor t, given and

Divide both sides by P and reverse the equation.Take the natural logarithm of both sides.

t L 10.66 years

t =

ln 20.065

0.065t = ln 2

ln e0.065t= ln 2

e0.065t= 2

2P = Pe0.065t

r = 0.065:A = 2PA = Pert,

MATCHED PROBLEM 4 How long will it take money to triple if it is invested at 5.5% compoundedcontinuously?

4. If $4,000 is invested at 8% compounded continuously,graph the amount in the account as a function of time fora period of 6 years.

B In Problems 5–10, solve for t or r to two decimal places.

5. 6. 7.

8. 9. 10. 3 = e10r2 = e5r3 = e0.25t

3 = e0.1r2 = e0.03t2 = e0.06t

Exercise 4-1

A Use a calculator to evaluate A to the nearest cent in Problems 1and 2.

1. and 8

2. and 10

3. If $6,000 is invested at 10% compounded continuously,graph the amount in the account as a function of time fora period of 8 years.

A = $5,000e0.08t for t = 1, 4,

A = $1,000e0.1t for t = 2, 5,



n

10 2.593 74

100

1,000

10,000

100,000

1,000,000

10,000,000

e = 2.718 281 828 459 Áq

pT

31 � 11>n24n

s

0.01 2.704 81

0.001

0.000 1

0.000 01

0 e = 2.718 281 828 459 Á

pp- 0.000 01

- 0.000 1

- 0.001

- 0.01

11 + s21>s

C In Problems 11 and 12, use a calculator to complete each tableto five decimal places.

11.

14. Use a calculator and a table of values to investigate

Do you think this limit exists? If so, what do you think it is?

15. It can be shown that the number e satisfies the inequality

Illustrate this condition by graphing

in the same viewing window, for

16. It can be shown that

for any real number s. Illustrate this equation graphicallyfor by graphing

in the same viewing window, for 1 … n … 50.

y2 = 7.389 056 099 L e2

y1 = 11 + 2>n2ns = 2

es= lim

n: q

a1 +

s

nbn

1 … n … 20.

y3 = 11 + 1>n2n + 1

y2 = 2.718 281 828 L e

y1 = 11 + 1>n2n

a1 +

1nbn

6 e 6 a1 +

1nbn + 1

n Ú 1

lims:0+

a1 +

1sb s

12.


Do you think this limit exists? If so, what do you think it is?

limn: q

11 + n21>n

Applications17. Continuous compound interest. Recently, Provident Bank

offered a 10-year CD that earns 5.51% compounded con-tinuously.

(A) If $10,000 is invested in this CD, how much will it beworth in 10 years?

(B) How long will it take for the account to be worth$15,000?

18. Continuous compound interest. Provident Bank also offersa 3-year CD that earns 5.28% compounded continuously.

(A) If $10,000 is invested in this CD, how much will it beworth in 3 years?

(B) How long will it take for the account to be worth$11,000?

19. Present value. A note will pay $20,000 at maturity10 years from now. How much should you be willing topay for the note now if money is worth 5.2% compound-ed continuously?

20. Present value. A note will pay $50,000 at maturity 5 yearsfrom now. How much should you be willing to pay for the note now if money is worth 6.4% compoundedcontinuously?

21. Continuous compound interest. An investor bought stockfor $20,000. Five years later, the stock was sold for$30,000. If interest is compounded continuously, whatannual nominal rate of interest did the original $20,000investment earn?

22. Continuous compound interest. A family paid $40,000cash for a house. Fifteen years later, the house was soldfor $100,000. If interest is compounded continuously, whatannual nominal rate of interest did the original $40,000investment earn?

23. Present value. Solving for P, we obtain

which is the present value of the amount A due in t yearsif money earns interest at an annual nominal rate r com-pounded continuously.

(A) Graph (B) [Guess, using part (A).]

[Conclusion: The longer the time until the amount A is due, the smaller is its present value, as we wouldexpect.]

limt: q 10,000e-0.08t= ?

P = 10,000e-0.08t, 0 … t … 50.

P = Ae-rt

A = Pert



24. Present value. Referring to Problem 23, in how manyyears will the $10,000 have to be due in order for itspresent value to be $5,000?

25. Doubling time. How long will it take money to double ifit is invested at 7% compounded continuously?

(C) Determine the doubling rates for 4, 6, 8, 10,and 12 years.

33. Radioactive decay. A mathematical model for the decayof radioactive substances is given by

where

If the continuous compound rate of decay of radium per year is how long will it take a certain amount of radium to decay to half the originalamount? (This period is the half-life of the substance.)

34. Radioactive decay. The continuous compound rate ofdecay of carbon-14 per year is Howlong will it take a certain amount of carbon-14 to decay to half the original amount? (Use the radioactive decaymodel in Problem 33.)

35. Radioactive decay. A cesium isotope has a half-life of 30 years. What is the continuous compound rate of decay?(Use the radioactive decay model in Problem 33.)

36. Radioactive decay. A strontium isotope has a half-life of90 years. What is the continuous compound rate of decay?(Use the radioactive decay model in Problem 33.)

37. World population. A mathematical model for worldpopulation growth over short intervals is given by

where

How long will it take world population to double if itcontinues to grow at its current continuous compoundrate of 1.3% per year?

38. U.S. population. How long will it take for the U.S. popula-tion to double if it continues to grow at a rate of 0.85%per year?

39. Population growth. Some underdeveloped nations have population doubling times of 50 years. At what continuous compound rate is the population growing? (Use the population growth model in Problem 37.)

40. Population growth. Some developed nations havepopulation doubling times of 200 years. At whatcontinuous compound rate is the population growing?(Use the population growth model in Problem 37.)

P = population at time t

t = time in years

r = continuous compound rate of growth

P0 = population at time t = 0

P = P0ert

r = -0.000 123 8.

r = -0.000 433 2,

Q = amount of the substance at time t

t = time in years

r = continuous compound rate of decay

Q0 = amount of the substance at time t = 0

Q = Q0ert

t = 2,

26. Doubling time. How long will it take money to double ifit is invested at 5% compounded continuously?

27. Doubling rate. At what nominal rate compounded con-tinuously must money be invested to double in 8 years?

28. Doubling rate. At what nominal rate compounded con-tinuously must money be invested to double in 10 years?

29. Growth time. A man with $20,000 to invest decides to di-versify his investments by placing $10,000 in an accountthat earns 7.2% compounded continuously and $10,000 inan account that earns 8.4% compounded annually. Usegraphical approximation methods to determine how longit will take for his total investment in the two accounts togrow to $35,000.

30. Growth time. A woman invests $5,000 in an account thatearns 8.8% compounded continuously and $7,000 in anaccount that earns 9.6% compounded annually. Usegraphical approximation methods to determine how longit will take for her total investment in the two accounts togrow to $20,000.

31. Doubling times

(A) Show that the doubling time t (in years) at an annualrate r compounded continuously is given by

(B) Graph the doubling-time equation from part (A) forIs this restriction on r reasonable?

Explain.(C) Determine the doubling times (in years, to two

decimal places) for 10%, 15%, 20%, 25%, and 30%.

32. Doubling rates

(A) Show that the rate r that doubles an investment atcontinuously compounded interest in t years is given by

(B) Graph the doubling-rate equation from part (A) forIs this restriction on t reasonable?

Explain.1 … t … 20.

r =

ln 2t

r = 5%,

0.02 … r … 0.30.

t =

ln 2r



Section 4-2 DERIVATIVES OF EXPONENTIAL ANDLOGARITHMIC FUNCTIONS� The Derivative of � The Derivative of ln � Other Logarithmic and Exponential Functions� Exponential and Logarithmic Models

In this section, we obtain formulas for the derivatives of logarithmic and exponentialfunctions. A review of Sections 2–4 and 2–5 may prove helpful. In particular, recallthat is the exponential function with base and the inverse of thefunction is the natural logarithm function ln x. More generally, if b is a positive realnumber, then the exponential function with base b, and the logarithmicfunction with base b, are inverses of each other.

� The Derivative of ex

In the process of finding the derivative of we will use (without proof) the fact that

(1)

Explore & Discuss 1 Complete Table 1.

limh:0

eh

- 1h

= 1

ex,

logb xbxb Z 1,

exe L 2.718f1x2 = ex

xex

TABLE 1

h �0.1 �0.01 �0.001 0 0.001 0.01 0.1

eh � 1h

;:

Do your calculations make it reasonable to conclude that

Discuss.

We now apply the four-step process (Section 3-4) to the exponential function

Step 1. Find

See Section 2-4.

Step 2. Find

Factor out

Step 3. Find

Step 4. Find f¿1x2 = limh:0

f1x + h2 - f1x2h

.

f1x + h2 - f1x2h

=

ex1eh- 12

h= ex

a eh- 1hb

f1x + h2 - f1x2h

.

= ex1eh- 12

ex. f1x + h2 - f1x2 = exeh- ex

f1x + h2 - f1x2.f1x + h2 = ex + h

= exeh

f1x + h2.f1x2 = ex.

limh:0

eh

- 1h

= 1?


S e c t i o n 4 - 2 Derivatives of Exponential and Logarithmic Functions 225

Use the limit in (1).

Thus,

The derivative of the exponential function is the exponential function.

ddx

ex= ex

= ex # 1 = ex

= ex limh:0

a eh- 1hb

= limh:0

ex a eh

- 1hb

f¿1x2 = limh:0

f1x + h2 - f1x2h

E X A M P L E 1 Finding Derivatives Find for

(A)

(B)

SOLUTIONS (A)

(B)

Remember that e is a real number, so the power rule (Section 3–5) is used to findthe derivative of xe. The derivative of the exponential function , however, is .Note that is a constant, so its derivative is 0.e2

L 7.389exex

f¿1x2 = -7exe - 1+ 2ex

f¿1x2 = 5ex- 12x3

+ 9

f1x2 = -7xe+ 2ex

+ e2

f1x2 = 5ex- 3x4

+ 9x + 16

f¿1x2

MATCHED PROBLEM 1 Find for

(A)

(B)

CAUTION:

The power rule cannot be used to differentiate the exponential function. The powerrule applies to exponential forms where the exponent is a constant and the base is a variable. In the exponential form the base is a constant and the exponent is avariable.

ex,xn

ddx

ex= exd

dx ex

Z xex - 1

f1x2 = x7- x5

+ e3- x + ex

f1x2 = 4ex+ 8x2

+ 7x - 14

f¿1x2

� The Derivative of ln xWe summarize some important facts about logarithmic functions from Section 2–5:

SUMMARY Recall that the inverse of an exponential function is called a logarithmic function.For and

Logarithmic form Exponential form

Range: 10, q2Range: 1- q , q2Domain: 1- q , q2Domain: 10, q2

x = byis equivalent to y = logb x

b Z 1,b 7 0


We are now ready to use the definition of the derivative and the four-step processdiscussed in Section 3-4 to find a formula for the derivative of ln x. Later we will ex-tend this formula to include for any base b.

Step 1. Find

cannot be simplified.

Step 2. Find

Use

Step 3. Find

=

1x

lna1 +

hxbx>h

=

1x

cxh

lna1 +

hxb d

=

xx

# 1h

ln x + h

x

=

1h

ln x + h

x

f1x + h2 - f1x2

h=

ln1x + h2 - ln x

h

f1x + h2 - f1x2h

.

= ln x + h

x

ln A - ln B = ln AB

. f1x + h2 - f1x2 = ln1x + h2 - ln x

f1x + h2 - f1x2.ln1x + h2f1x + h2 = ln1x + h2

f1x + h2.Let f1x2 = ln x, x 7 0.

logb x


x

y

5�5

5

10

10

�5

y � bx

y � x

y � logb x

FIGURE 1

The graphs of and are symmetric with respect to the line (See Figure 1.)

y = x.y = bxy = logb x

Of all the possible bases for logarithmic functions, the two most widely used are

Common logarithm (base 10)Natural logarithm (base e) ln x = loge x

log x = log10 x

Multiply by to change form.

Use p ln A = ln Ap.

1 = x>x

Step 4. Find

= limh:0

c 1x

lna1 +

hxbx>h d

f¿1x2 = limh:0

f1x + h2 - f1x2h

f¿1x2 = limh:0

f1x + h2 - f1x2h

.

Let Note that implies s : 0.

h : 0s = h>x.



=

1x

=

1x

ln e

=

1x

ln c lims:011 + s21>s d

=

1x

lims:0

3ln11 + s21>s4 Use a new limit property.*

Use the definition of e.

ln e = loge e = 1

Thus,d

dx ln x =

1x

In the derivation of the derivative of ln x, we used the following properties of logarithms:

We also noted that there is no property that simplifies (See Theorem 1 inSection 2-5 for a list of properties of logarithms.)

ln1A + B2.ln

A

B= ln A - ln B ln Ap

= p ln A

I N S I G H T


(A)

(B)

SOLUTIONS (A)

(B) Before taking the derivative, we use a property of logarithms (see Theorem 1,Section 2-5) to rewrite y.

Use ln Mp � p ln M.Now take the derivative of both sides.

y¿ = 4x3-

4x

y = x4- 4 ln x

y = x4- ln x4

y¿ = 3ex+

5x

y = x4- ln x4

y = 3ex+ 5 ln x

y¿


(A)(B) y = ln x5

+ ex- ln e2

y = 10x3- 100 ln x

y¿

* The following new limit property is used: If exists and is positive, then ln3limx:c f1x24.

limx:c3ln f1x24 =limx:c f1x2

� Other Logarithmic and Exponential FunctionsIn most applications involving logarithmic or exponential functions, the number e isthe preferred base. However, in some situations it is convenient to use a base otherthan e. Derivatives of and can be obtained by expressing thesefunctions in terms of the natural logarithmic and exponential functions.

y = bxy = logb x


We begin by finding a relationship between and ln x for any base and Some of you may prefer to remember the process, others the formula.

y =

1ln b

ln x

y ln b = ln x

ln by= ln x

by= x

y = logb x

b Z 1.b, b 7 0logb x


Change to exponential form.Take the natural logarithm of both sides.Recall that Solve for y.

ln by= y ln b.

Thus,

Change-of-base formula for logarithms* (2)

Similarly, we can find a relationship between bx and ex for any base

Take the natural logarithm of both sides.Recall that ln bx � x ln b.Take the exponential function of both sides.

Thus,

Change-of-base formula for exponential functions (3)

Differentiating both sides of equation (2) gives

It can be shown that the derivative of the function ecx, where c is a constant, is thefunction cecx (see Problems 49–50 in Exercise 4-2 or the more general results of Sec-tion 4–4). Therefore, differentiating both sides of equation (3), we have

For convenient reference, we list the derivative formulas that we have obtained forexponential and logarithmic functions:

d

dx bx � ex ln b ln b � bx ln b

d

dx logb x �

1ln b

d

dx ln x �

1ln b

a1xb

bx= ex ln b

y = ex ln b

ln y = x ln b

ln y = ln bx

y = bx

b, b 7 0, b Z 1.

logb x =

1ln b

ln x

* Equation (2) is a special case of the general change-of-base formula for logarithms (which can be derivedin the same way): logb x = (loga x)�(loga b).

DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

For

d

dx logb x =

1ln b

a 1xbd

dx ln x =

1x

d

dx bx

= bx ln bd

dx ex

= ex

b 7 0, b Z 1,


(A)

(B) g1x2 = log4 x5

g1x2 = 2x- 3x

g¿1x2



SOLUTIONS (A)

(B) First use a property of logarithms to rewrite g(x).

Use .Take the derivative of both sides.

g¿1x2 =

5ln 4

a 1xb

g1x2 = 5 log4 x

logb Mp= p logb M g1x2 = log4 x

5

g¿1x2 = 2x ln 2 - 3x ln 3


(A)

(B) g1x2 = log2 x - 6 log5 x

g1x2 = x10+ 10x

g¿1x2

Explore & Discuss 2 (A) The graphs of and are shown in Figure 2.Which graph belongs to which function?

g(x) = log4 xf(x) = log2 x

x

y

108642

4

3

2

1

FIGURE 2

(B) Sketch graphs of and

(C) The function f(x) is related to g(x) in the same way that is related toWhat is that relationship?g¿(x).

f ¿(x)

g¿(x).f ¿(x)

� Exponential and Logarithmic Models

E X A M P L E 4 Price–Demand Model An Internet store sells blankets made of the finestAustralian wool. If the store sells x blankets at a price of $p per blanket, then theprice–demand equation is . Find the rate of change of price withrespect to demand when the demand is 800 blankets, and interpret the result.

SOLUTION

If then

When the demand is 800 blankets, the price is decreasing about $0.16 per blanket.

dp

dx= 35010.9992800 ln 0.999 L -0.157, or - $0.16

x = 800,

dp

dx= 35010.9992x ln 0.999

p = 35010.9992x

MATCHED PROBLEM 4 The store in Example 4 also sells a reversible fleece blanket. If the price–demand equa-tion for reversible fleece blankets is find the rate of change of pricewith respect to demand when the demand is 400 blankets and interpret the result.

p = 20010.9982x,


Answers to Matched Problems 1. (A)

(B)

2. (A)

(B)

3. (A)

(B)

4. The price is decreasing at the rate of $0.18 per blanket.

5. The circulation in 2010 is approximately 52.4 million and is decreasing at the rate of 0.3million per year.

a 1ln 2

-

6ln 5b

1x

10x9+ 10x ln 10

5x

+ ex

30x2-

100x

7x6- 5x4

- 1 + ex

4ex+ 16x + 7


E X A M P L E 5 Cable TV Subscribers A statistician used data from the U.S. census to constructthe model

where S(t) is the number of cable TV subscribers (in millions) in year t ( corre-sponds to 1980). Use this model to estimate the number of cable TV subscribers in2010 and the rate of change of the number of subscribers in 2010 (round both to thenearest tenth of a million). Interpret these results.

SOLUTION Since 2010 corresponds to we must find S(30) and

In 2010 there will be approximately 73.4 million subscribers, and this number is grow-ing at the rate of 0.7 million per year.

S¿1302 =

2130

= 0.7 million

S¿1t2 = 21 1t

=

21t

S1302 = 21 ln 30 + 2 = 73.4 million

S¿1302.t = 30,

t = 0

S1t2 = 21 ln t + 2

MATCHED PROBLEM 5 A model for newspaper circulation is

where C(t) is newspaper circulation (in millions) in year t ( corresponds to 1980).Use this model to estimate the circulation and the rate of change of circulation in2010 (round both to the nearest tenth of a million). Interpret these results.

t = 0

C(t) = 83 - 9 ln t

On most graphing calculators, exponential regression produces a function of the formFormula (3) enables you to change the base b (chosen by the graphing cal-

culator) to the more familiar base e:

On most graphing calculators, logarithmic regression produces a function of the formFormula (2) enables you to write the function in terms of logarithms to

any base d that you may prefer:

y = a + b ln x = a + b1ln d2 log d x

y = a + b ln x.

y = a # bx= a # ex ln b

y = a # bx.

I N S I G H T



Exercise 4-2

A In Problems 1–14, find

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

B In Problems 15–22, find the equation of the line tangent to thegraph of f at the indicated value of x.

15.

16.

17.

18.

19.

20.

21.

22.

23. A student claims that the line tangent to the graph ofat passes through the point (2, 0)

(see the figure). Is she correct? Will the line tangent atpass through (3, 0)? Explain.x = 4

x = 3f1x2 = ex

f1x2 = 5ex; x = 1

f1x2 = 2 + ex; x = 1

f1x2 = 1 + ln x4; x = e

f1x2 = ln x3; x = e

f1x2 = ex+ 1; x = 0

f1x2 = 3ex; x = 0

f1x2 = 2 ln x; x = 1

f1x2 = 3 + ln x; x = 1

f1x2 = ln x10+ 2 ln x

f1x2 = ln x2+ 4ex

f1x2 = 4 + ln x9

f1x2 = 5x - ln x5

f1x2 = ln x 8

f1x2 = ln x3

f1x2 = ln x + 2ex- 3x2

f1x2 = ex+ x - ln x

f1x2 = 9ex+ 2x2

f1x2 = x3- 6ex

f1x2 = 6 ln x - x3+ 2

f1x2 = -2 ln x + x2- 4

f1x2 = -7ex- 2x + 5

f1x2 = 5ex+ 3x + 1

f¿1x2. 25. A student claims that the line tangent to the graph ofpasses through the origin (see the

figure). Is he correct? Will the line tangent at passthrough the origin? Explain.

x = 4g(x) = ln x at x = 3

x431 2�1

f (x)

�10

40

30

20

10

Figure for 23

x632 41 5�1

g(x)

�1

2

1

Figure for 25

26. Refer to Problem 25. Does the line tangent to the graphof at pass through the origin? Are thereany other lines tangent to the graph of f that pass throughthe origin? Explain.

In Problems 27–30, first use appropriate properties of loga-rithms to rewrite f(x), and then find

27.

28.

29.

30.

C In Problems 31–42, find for the indicated function y.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

In Problems 43–48, use graphical approximation methods tofind the points of intersection of f(x) and g(x) (to two decimalplaces).

43.[Note that there are three points of intersection and that

is greater than for large values of x.]x4ex

f1x2 = ex; g1x2 = x4

y = e3- 3x

y = 2x+ e2

y = - log2 x + 10 ln x

y = 3 ln x + 2 log3 x

y = x5- 5x

y = 10 + x + 10x

y = log x + 4x2+ 1

y = 2x - log x

y = 4x

y = 3x

y = 3 log5 x

y = log 2 x

dy

dx

f1x2 = x + 5 ln 6x

f1x2 = ln 4

x3

f1x2 = 2 + 3 ln 1x

f1x2 = 10x + ln 10x

f¿1x2.

x = ef(x) = ln x

24. Refer to Problem 23. Does the line tangent to the graphof at pass through the origin? Are thereany other lines tangent to the graph of f that pass throughthe origin? Explain.

x = 1f1x2 = ex


44.

[Note that there are two points of intersection and that is greater than for large values of x.]

45.

46. f1x2 = 1ln x23; g1x2 = 2x

f1x2 = 1ln x22; g1x2 = x

x5ex

f1x2 = ex; g1x2 = x5


47.

48.

49. Explain why .

50. Use the result of Problem 49 and the four-step process to show that if f1x2 = ecx, then f¿1x2 = cecx.

limh:0

ech

- 1h

= c

f1x2 = ln x; g1x2 = x1>4f1x2 = ln x; g1x2 = x1>5

Applications51. Salvage value. The salvage value S (in dollars) of a

company airplane after t years is estimated to be given by

What is the rate of depreciation (in dollars per year) after1 year? 5 years? 10 years?

S1t2 = 300,00010.92tnormal children. Using hospital records for 5,000 normalchildren, the experimenters found that the systolic bloodpressure was given approximately by

where P(x) is measured in millimeters of mercury and x is measured in pounds. What is the rate of change of blood pressure with respect to weight at the 40-pound weight level? At the 90-pound weight level?

56. Refer to Problem 55. Find the weight (to the nearestpound) at which the rate of change of blood pressure with respect to weight is 0.3 millimeter of mercury per pound.

57. Psychology: stimulus/response. In psychology, theWeber–Fechner law for the response to a stimulus is

where R is the response, S is the stimulus, and is the lowest level of stimulus that can be detected. Find .

58. Psychology: learning. A mathematical model for theaverage of a group of people learning to type is given by

where N(t) is the number of words per minute typed aftert hours of instruction and practice (2 hours per day, 5 daysper week). What is the rate of learning after 10 hours ofinstruction and practice? After 100 hours?

N1t2 = 10 + 6 ln t t Ú 1

dR>dS

S0

R = k ln S

S0

P(x) = 17.5(1 + ln x) 10 � x � 100

52. Resale value. The resale value R (in dollars) of a compa-ny car after t years is estimated to be given by

What is the rate of depreciation (in dollars per year) after1 year? 2 years? 3 years?

53. Bacterial growth. A single cholera bacterium dividesevery 0.5 hour to produce two complete cholera bacteria.If we start with a colony of 5,000 bacteria, after t hoursthere will be

bacteria. Find and and interpret theresults.

54. Bacterial growth. Repeat Problem 53 for a startingcolony of 1,000 bacteria such that a single bacteriumdivides every 0.25 hour.

55. Blood pressure. An experiment was set up to find a rela-tionship between weight and systolic blood pressure in

A¿(5),A¿(t), A¿(1),

A1t2 = 5,000 # 22t= 5,000 # 4t

R1t2 = 20,00010.862t

Section 4-3 DERIVATIVES OF PRODUCTS AND QUOTIENTS� Derivatives of Products� Derivatives of Quotients

The derivative properties discussed in Section 3-5 added substantially to our abilityto compute and apply derivatives to many practical problems. In this and the next twosections, we add a few more properties that will increase this ability even further.

� Derivatives of ProductsIn Section 3-5, we found that the derivative of a sum is the sum of the derivatives. Isthe derivative of a product the product of the derivatives?


S e c t i o n 4 - 3 Derivatives of Products and Quotients 233

Explore & Discuss 1 Let and Which of the followingis

(A)

(B)

(C)

(D)

Comparing the various expressions computed in Explore & Discuss 1, we see thatthe derivative of a product is not the product of the derivatives, but appears to involvea slightly more complicated form.

Using the definition of the derivative and the four-step process, we can show that

The derivative of the product of two functions is the first function times thederivative of the second function, plus the second function times the derivativeof the first function.

That is,

F1x2S¿1x2 + F¿1x2S1x2F¿1x2S1x2F1x2S¿1x2F¿1x2S¿1x2

f¿1x2? f1x2 = F1x2S1x2 = x5.F1x2 = x2, S1x2 = x3,

THEOREM 1 PRODUCT RULEIf

and if and exist, then

Also,

y¿ = FS¿ + SF¿ dy

dx= F

dS

dx+ S

dF

dx

f¿1x2 = F1x2S¿1x2 + S1x2F¿1x2S¿1x2F¿1x2

y = f1x2 = F1x2S1x2

E X A M P L E 1 Differentiating a Product Use two different methods to find for

SOLUTION Method 1. Use the product rule:

= 36x5- 8x

= 24x5+ 12x5

- 8x

= 2x2112x32 + 13x4- 2214x2

f ¿1x2 = 2x213x4- 22¿ + 13x4

- 2212x22¿

f1x2 = 2x213x4- 22.

f¿1x2

First times derivative ofsecond, plus second timesderivative of first

Method 2. Multiply first; then take derivatives:

f¿1x2 = 36x5- 8x

f1x2 = 2x213x4- 22 = 6x6

- 4x2

MATCHED PROBLEM 1 Use two different methods to find for f1x2 = 3x312x2- 3x + 12.f¿1x2

Some of the products we encounter can be differentiated by either of the methodsillustrated in Example 1. In other situations, the product rule must be used. Unlessinstructed otherwise, you should use the product rule to differentiate all products inthis section in order to gain experience with the use of this important differentiationrule.



E X A M P L E 2 Tangent Lines Let

(A) Find the equation of the line tangent to the graph of f(x) at

(B) Find the value(s) of x where the tangent line is horizontal.

SOLUTION (A) First, find

Then, find f (3) and

Now, find the equation of the tangent line at

Tangent line at

(B) The tangent line is horizontal at any value of x such that so

The tangent line is horizontal at and at x = 2.x = 1

x = 1, 2

1x - 121x - 22 = 0

x2- 3x + 2 = 0

6x2- 18x + 12 = 0

f¿1x2 = 12x - 922x + 1x2+ 622 = 0

f¿1x2 = 0,

x = 3 y = 12x - 81

m = f ¿1x12 = f ¿132 = 12 y - 1-452 = 121x - 32y1 = f1x12 = f132 = -45 y - y1 = m1x - x12

x = 3:

f¿132 = 32132 - 942132 + 132+ 62122 = -18 + 30 = 12

f132 = 32132 - 94132+ 62 = 1-321152 = -45

f¿132: = 12x - 9212x2 + 1x2

+ 62122 f¿1x2 = 12x - 921x2

+ 62¿ + 1x2+ 6212x - 92¿

f¿1x2:

x = 3.

f1x2 = 12x - 921x2+ 62.

MATCHED PROBLEM 2 Repeat Example 2 for f1x2 = 12x + 921x2- 122.

As Example 2 illustrates, the way we write depends on what we want to do with it.If we are interested only in evaluating at specified values of x, the form in part (A)is sufficient. However, if we want to solve we must multiply and collect liketerms, as we did in part (B).

f¿1x2 = 0,f¿1x2

f¿1x2I N S I G H T


(A)

(B)

SOLUTIONS

(A)

(B)

= 6x3 11 + 4 ln x2 = 6x3

+ 24x3 ln x

= 6x4 1x

+ 1ln x2124x32 f¿1x2 = 6x4 1ln x2¿ + 1ln x2 16x42¿

= 2x2ex1x + 32 = 2x3ex

+ ex16x22 f¿1x2 = 2x31ex2¿ + ex12x32¿

f1x2 = 6x4 ln x

f1x2 = 2x3ex

f¿1x2



� Derivatives of QuotientsAs is the case with a product, the derivative of a quotient of two functions is not thequotient of the derivatives of the two functions.

Explore & Discuss 2 Let and

Which of the following is

(A) (B) (C)

(D)

The expressions in Explore & Discuss 2 suggest that the derivative of a quotientleads to a more complicated quotient than you might expect.

In general, if T(x) and B(x) are any two differentiable functions and

then

Thus,

The derivative of the quotient of two functions is the denominator functiontimes the derivative of the numerator function, minus the numerator functiontimes the derivative of the denominator function, divided by the denominatorfunction squared.

f¿1x2 =

B1x2T¿1x2 - T1x2B¿1x23B1x242

f1x2 =

T1x2B1x2

T¿1x2B1x23B1x242 -

T1x2B¿1x23B1x242 =

T¿1x2B1x2 - T1x2B¿1x23B1x242

T1x2B¿1x23B1x242

T¿1x2B1x23B1x242

T¿1x2B¿1x2

f¿1x2?f1x2 =

T1x2B1x2 =

x5

x2 = x3

T1x2 = x5, B1x2 = x2,


(A)

(B) f1x2 = x7 ln x

f1x2 = 5x8 ex

f ¿1x2

THEOREM 2 QUOTIENT RULEIf

and if and exist, then

Also,

y¿ =

BT¿ - TB¿

B2 dy

dx=

B dT

dx- T

dB

dx

B2

f¿1x2 =

B1x2T¿1x2 - T1x2B¿1x23B1x242

B¿1x2T¿1x2y = f1x2 =

T1x2B1x2


SOLUTION (A)

=

2x2- 2x

12x - 122

=

4x2- 2x - 2x2

12x - 122

=

12x - 1212x2 - x212212x - 122

f¿1x2 =

12x - 121x22¿ - x212x - 12¿12x - 122


E X A M P L E 4 Differentiating Quotients

(A) If (B) If

(C) Find by using the quotient rule and also by splitting the fraction

into two fractions.

d

dx x2

- 3

x2

y =

t2- t

t3+ 1

, find y¿.f1x2 =

x2

2x - 1, find f¿1x2.

The denominator times the derivative ofthe numerator, minus the numeratortimes the derivative of the denominator,divided by the square of the denominator

(B)

(C) Method 1. Use the quotient rule:

Method 2. Split into two fractions:

Comparing methods 1 and 2, we see that it often pays to change an expression alge-braically before blindly using a differentiation formula.

d

dx 11 - 3x-22 = 0 - 31-22x-3

=

6

x3

x2- 3

x2 =

x2

x2 -

3

x2 = 1 - 3x-2

=

2x3- 2x3

+ 6x

x4 =

6x

x4 =

6

x3

=

x212x2 - 1x2- 322x

x4

d

dx x2

- 3

x2 =

x2 d

dx 1x2

- 32 - 1x2- 32

d

dx x2

1x222

=

- t4+ 2t3

+ 2t - 1

1t3+ 122

=

2t4- t3

+ 2t - 1 - 3t4+ 3t3

1t3+ 122

=

1t3+ 1212t - 12 - 1t2

- t213t221t3

+ 122

y¿ =

1t3+ 121t2

- t2¿ - 1t2- t21t3

+ 12¿1t3

+ 122

MATCHED PROBLEM 4 Find

(A)

(B)

(C) in two waysd

dx 2 + x3

x3

y¿ for y =

t3- 3t

t2- 4

f¿1x2 for f1x2 =

2x

x2+ 3



Explore & Discuss 3 Explain why is used in the following expression, and then find the correctderivative:

d

dx

x3

x2+ 3x + 4

Z

3x2

2x + 3

Z


(A)

(B)

SOLUTIONS (A)

(B)

Multiply by

=

2x + 5 - 2x ln x

x12x + 522

xx

=

12x + 52 # 1x

- 1ln x2212x + 522

f¿1x2 =

12x + 521ln x2¿ - 1ln x212x + 52¿12x + 522

=

3ex

11 + ex22

=

11 + ex23ex- 3exex

11 + ex22

f¿1x2 =

11 + ex213ex2¿ - 3ex11 + ex2¿11 + ex22

f1x2 =

ln x2x + 5

f1x2 =

3ex

1 + ex

f¿1x2


(A)

(B) f1x2 =

4x

1 + ln x

f¿1x2 =

x3

ex+ 2

f¿1x2

E X A M P L E 6 Sales Analysis The total sales S (in thousands of games) of a home video game tmonths after the game is introduced are given by

(A) Find

(B) Find S(10) and Write a brief verbal interpretation of these results.

(C) Use the results from part (B) to estimate the total sales after 11 months.

S¿(10).

S¿(t).

S(t) =

125t2

t2+ 100


SOLUTION (A)

(B)

The total sales after 10 months are 62,500 games, and sales are increasing at therate of 6,250 games per month.

(C) The total sales will increase by approximately 6,250 games during the next month.Thus, the estimated total sales after 11 months are games.

62,500 + 6,250 = 68,750

S1102 =

12511022102

+ 100= 62.5 and S¿1102 =

25,00011021102

+ 10022 = 6.25

=

25,000t

1t2+ 10022

=

250t3+ 25,000t - 250t3

1t2+ 10022

=

1t2+ 10021250t2 - 125t212t2

1t2+ 10022

S¿1t2 =

1t2+ 10021125t22¿ - 125t21t2

+ 1002¿1t2

+ 10022


MATCHED PROBLEM 6 Refer to Example 6. Suppose that the total sales S (in thousands of games) t months after the game is introduced are given by

(A) Find (B) Find S(12) and Write a brief verbal interpretation of these results.(C) Use the results from part (B) to estimate the total sales after 13 months.

S¿1122.S¿1t2.

S1t2 =

150t

t + 3

Answers to Matched Problems 1.

2. (A)

(B)

3. (A)

(B)

4. (A)

(B)

(C)

5. (A)

(B)

6. (A)

(B) After 12 months, the total sales are 120,000 games, and salesare increasing at the rate of 2,000 games per month.

(C) 122,000 games

S1122 = 120; S¿1122 = 2.

S¿1t2 =

450

1t + 322

11 + ln x2 4 - 4x 1x

11 + ln x22 =

4 ln x

11 + ln x22

1ex+ 22 3x2

- x3 ex

1ex+ 222

-

6

x4

1t2- 4213t2

- 32 - 1t3- 3t212t2

1t2- 422 =

t4- 9t2

+ 12

1t2- 422

1x2+ 322 - 12x212x21x2

+ 322 =

6 - 2x2

1x2+ 322

x7 # 1x

+ ln x 17x62 = x6 11 + 7 ln x25x8ex

+ ex140 x72 = 5x71x + 82ex

x = -4, x = 1

y = 84x - 297

30x4- 36x3

+ 9x2

BARNMC04_0132328186.QXD 2/21/07 1:27 PM 8


The answers to most of the problems in this exercise set containboth an unsimplified form and a simplified form of the deriva-tive. When checking your work, first check that you applied therules correctly, and then check that you performed the algebraicsimplification correctly. Unless instructed otherwise, when dif-ferentiating a product, use the product rule rather than per-forming the multiplication first.

A In Problems 1–26, find and simplify.

1. 2.

3.

4. 5.

6. 7.

8. 9.

10. 11.

12.

13.

14.

15.

16.

17. 18.

19.

20.

21. 22.

23. 24.

25. 26.

In Problems 27–38, find where f(x) is an unspecifieddifferentiable function.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

B In Problems 39–48, find the indicated derivatives and simplify.

39.

40. y¿ for y = 1x3+ 2x2213x - 12

f¿1x2 for f1x2 = 12x + 121x2- 3x2

h1x2 =

f1x2ln x

h1x2 =

ln xf1x2

h1x2 =

ex

f1x2h1x2 = exf1x2

h1x2 =

x2

f1x2h1x2 =

x

f1x2

h1x2 =

f1x2x3h1x2 =

f1x2x2

h1x2 =

f1x2x

h1x2 = x3f1x2h1x2 = x2f1x2h1x2 = xf1x2

h¿1x2,f1x2 =

2x

1 + ln xf1x2 =

ln x1 + x

f1x2 =

1 - ex

1 + exf1x2 =

ex

x2+ 1

f1x2 =

x2- 4

x2+ 5

f1x2 =

x2+ 2

x2- 3

f1x2 = 1x2- 421x2

+ 52f1x2 = 1x2

+ 221x2- 32

f1x2 =

3x + 5

x2- 3

f1x2 =

x2+ 1

2x - 3

f1x2 = 10.5x - 4210.2x + 12f1x2 = 10.4x + 2210.5x - 52f1x2 = 13x + 521x2

- 32f1x2 = 1x2

+ 1212x - 32f1x2 = 5x ln x

f1x2 = x3 ln xf1x2 = x2ex

f1x2 = 3xexf1x2 =

3x - 42x + 3

f1x2 =

2x + 3x - 2

f1x2 =

3x

2x + 1

f1x2 =

x

x - 3f1x2 = 13x + 2214x - 52f1x2 = 1x - 3212x - 12

f1x2 = 5x21x3+ 22f1x2 = 2x31x2

- 22f¿1x2

41.

42.

43.

44.

45.

46.

47.

48.

In Problems 49–54, find and find the equation of the line tangent to the graph of f at

49.

50.

51. 52.

53. 54.

In Problems 55–58, find and find the value(s) of xwhere

55.

56.

57. 58.

In Problems 59–62, find in two ways: by using theproduct or quotient rule and by simplifying first.

59. 60.

61. 62.

C In Problems 63–82, find each indicated derivative and simplify.

63.

64. 65.

66.

67.

68.

69. 70.

71. f¿1x2 for f1x2 =

6 32x

x2- 3

dy

dx for y =

10x

1 + x4y¿ for y =

log2 x

1 + x2

d

dx 314x1>2

- 1213x1>3+ 224

dy

dx for y = 9x1>31x3

+ 52y¿ for y =

x3- 3x + 4

2x2+ 3x - 2

d

dx 3x2

- 2x + 3

4x2+ 5x - 1

g1w2 = 1w - 52 log3 w

f1w2 = 1w + 122w

f1x2 =

x4+ 4

x4f1x2 =

x3+ 9

x3

f1x2 = x41x3- 12f1x2 = x31x4

- 12

f¿1x2f1x2 =

x

x2+ 9

f1x2 =

x

x2+ 1

f1x2 = 12x - 321x2- 62

f1x2 = 12x - 1521x2+ 182

f¿1x2 = 0.f¿1x2

f1x2 = 1x - 22 ln xf1x2 =

x

2x

f1x2 =

2x - 52x - 3

f1x2 =

x - 83x - 4

f1x2 = 17 - 3x211 + 2x2f1x2 = 11 + 3x215 - 2x2

x = 2.f¿1x2

dy

dt for y = 11 + et2 ln t

y¿ for y = 11 + x - x22 ex

dy

dw for y =

w4- w3

3w - 1

d

dw w2

- 3w + 1

w2- 1

f¿1x2 for f1x2 =

3x2

2x - 1

y¿ for y =

5x - 3

x2+ 2x

d

dt 313 - 0.4t3210.5t2

- 2t24

dy

dt for y = 12.5t - t2214t + 1.42

Exercise 4-3


72.

73.

74. 75.

76. 77.d

dx x3

- 2x2

32x2

d

dt [10 t log t]

d

dx [4x log x5]h¿1t2 if h1t2 =

-0.05t2

2t + 1

g¿1t2 if g1t2 =

0.2t

3t2- 1

y¿ for y =

22x

x2- 3x + 1

78.

79.

80.

81. 82.dy

du for y =

u2 eu

1 + ln u

dy

dt for y =

t ln tet

y¿ for y =

2x - 1

1x3+ 221x2

- 32

f¿1x2 for f1x2 =

12x2- 121x2

+ 32x2

+ 1

dy

dx for y =

x2- 3x + 1

42x


83. Sales analysis. The total sales S (in thousands of CDs) of a compact disk are given by

where t is the number of months since the release of the CD.

(A) Find (B) Find S(10) and Write a brief verbal inter-

pretation of these results.(C) Use the results from part (B) to estimate the total

sales after 11 months.

84. Sales analysis. A communications company has installeda cable television system in a city. The total number N (inthousands) of subscribers t months after the installationof the system is given by

(A) Find (B) Find N(16) and Write a brief verbal

interpretation of these results.(C) Use the results from part (B) to estimate the total

number of subscribers after 17 months.

85. Price–demand equation. According to classical economictheory, the demand x for a quantity in a free market de-creases as the price p increases (see the figure). Supposethat the number x of CD players people are willing to buyper week from a retail chain at a price of $p is given by

x =

4,0000.1p + 1

10 … p … 70

N¿1162.N¿1t2.

N1t2 =

180t

t + 4

S¿1102.S¿1t2.

S1t2 =

90t2

t2+ 50

price is $40. Write a brief verbal interpretation ofthese results.

(C) Use the results from part (B) to estimate the demandif the price is increased to $41.

86. Price–supply equation. Also according to classical eco-nomic theory, the supply x of a quantity in a free marketincreases as the price p increases (see the figure). Supposethat the number x of CD players a retail chain is willing tosell per week at a price of $p is given by

(A) Find dx�dp.

(B) Find the supply and the instantaneous rate of changeof supply with respect to price when the price is $40.Write a brief verbal interpretation of these results.

(C) Use the results from part (B) to estimate the supplyif the price is increased to $41.

87. Medicine. A drug is injected into the bloodstream of apatient through her right arm. The concentration of thedrug (in milligrams per cubic centimeter) in the blood-stream of the left arm t hours after the injection is given by

(A) Find

(B) Find and and interpret the results.

88. Drug sensitivity. One hour after a dose of x milligrams ofa particular drug is administered to a person, the changein body temperature T(x), in degrees Fahrenheit, is givenapproximately by

The rate at which T changes with respect to the size ofthe dosage, x, is called the sensitivity of the body to the dosage.

(A) Use the product rule to find .

(B) Find and

89. Learning. In the early days of quantitative learningtheory (around 1917), L. L. Thurstone found that a givenperson successfully accomplished N(x) acts after xpractice acts, as given by

(A) Find the instantaneous rate of change of learning,with respect to the number of practice acts, x.

(B) Find and N¿1682.N¿142N¿1x2,

N1x2 =

100x + 200x + 32

T¿162.T¿112, T¿132,T¿1x2

T¿1x2T1x2 = x2

a1 -

x

9b 0 … x … 7

C¿132,C¿10.52C¿1t2.

C1t2 =

0.14t

t2+ 1

x =

100p

0.1p + 1 10 … p … 70

Applications

x

p20 8040 60

Price (dollars)

Supp

ly/D

eman

d

400

800

1,200

2,000

1,600

x � 0.1p � 14,000

x � 0.1p � 1100p

Figure for 85 and 86

(A) Find dx�dp.

(B) Find the demand and the instantaneous rate ofchange of demand with respect to price when the


Section 4-4 THE CHAIN RULE� Composite Functions� General Power Rule� The Chain Rule

The word chain in the name “chain rule” comes from the fact that a function formedby composition involves a chain of functions—that is, a function of a function. Thechain rule enables us to compute the derivative of a composite function in terms ofthe derivatives of the functions making up the composition. In this section, we reviewcomposite functions, introduce the chain rule by means of the special case known asthe general power rule, and then discuss the chain rule itself.

� Composite FunctionsThe function is a combination of a quadratic function and a cubicfunction. To see this more clearly, let

Then we can express y as a function of x as follows:

The function m is said to be the composite of the two functions f and g. In general,we have the following:

y = f1u2 = f3g1x24 = 3x2+ 443 = m1x2

y = f1u2 = u3 and u = g1x2 = x2+ 4

m1x2 = 1x2+ 423

S e c t i o n 4 - 4 The Chain Rule 241

DEFINITION Composite Functions

A function m is a composite of functions f and g if

The domain of m is the set of all numbers x such that x is in the domain of g and g(x)is in the domain of f.

m1x2 = f3g1x24

E X A M P L E 1 Composite Functions Let and Find f [g(x)] and g[ f(u)].

SOLUTION

g3f1u24 = g1eu2 = -3eu

f 3g1x24 = f1-3x2 = e-3x

g1x2 = -3x.f1u2 = eu

MATCHED PROBLEM 1 Let and Find f[g(x)] and g[ f(u)].g1x2 = ex.f1u2 = 2u

E X A M P L E 2 Composite Functions Write each function as a composition of two simplerfunctions.

(A)

(B)

SOLUTION (A) Let

Check: y = f3g1x24 = 100eg1x2= 100e0.04x

u = g1x2 = 0.04x

y = f1u2 = 100eu

y = 24 - x2

y = 100e0.04x


(B) Let

Check: y = f3g1x24 = 2g1x2 = 24 - x2

u = g1x2 = 4 - x2

y = f1u2 = 2u


MATCHED PROBLEM 2 Write each function as a composition of two simpler functions.

(A) (B) y = 23 1 + x3y = 50e-2x

There can be more than one way to express a function as a composition of simpler func-tions. Choosing and in Example 2A produces thesame result:

Since we will be using composition as a means to an end (finding a derivative), usually itwill not matter what choices you make for the functions in the composition.

y = f3g1x24 = 100g1x2 = 100e0.04x

u = g1x2 = e0.04xy = f1u2 = 100u

I N S I G H T

� General Power RuleWe have already made extensive use of the power rule,

(1)

Now we want to generalize this rule so that we can differentiate composite functionsof the form where u(x) is a differentiable function. Is rule (1) still valid if wereplace x with a function u(x)?

Explore & Discuss 1 Let and Which of the following is

(A)

(B)

(C)

The calculations in Explore & Discuss 1 show that we cannot generalize the powerrule simply by replacing x with u(x) in equation (1).

How can we find a formula for the derivative of where u(x) is an arbi-trary differentiable function? Let’s begin by considering the derivatives of and to see if a general pattern emerges. Since we usethe product rule to write

(2)

Because we now use the product rule and the result inequation (2) to write

= 33u1x242u¿1x2 = 3u1x242u¿1x2 + u1x232u1x2u¿1x24 = 3u1x242

d

dx u1x2 + u1x2

d

dx 3u1x242

d

dx 3u1x243 =

d

dx 53u1x242u1x26

3u1x243 = 3u1x242u1x2, = 2u1x2u¿1x2 = u1x2u¿1x2 + u1x2u¿1x2

d

dx 3u1x242 =

d

dx 3u1x2u1x24

3u1x242 = u1x2u1x2,3u1x243 3u1x2423u1x24n,

33u1x242u¿1x233u¿1x24233u1x242

f¿1x2?f1x2 = 3u1x243 = 8x6.u1x2 = 2x2

3u1x24n,

d

dx xn

= nxn - 1

Use equation (2) tosubstitute for ddx

3u1x242.


Continuing in this fashion, we can show that

(3)

Using more advanced techniques, we can establish formula (3) for all real numbers n.Thus, we have the general power rule:

d

dx 3u1x24n = n3u1x24n - 1u¿1x2 n a positive integer


THEOREM 1 GENERAL POWER RULEIf u(x) is a differentiable function, n is any real number, and

then

This rule is often written more compactly as

y¿ = nun - 1u¿ or d

dx un

= nun - 1 du

dx where u = u1x2

f¿1x2 = n3u1x24n - 1u¿1x2

y = f1x2 = 3u1x24n

E X A M P L E 3 Using the General Power Rule Find the indicated derivatives:

(A) if

(B) if

(C)

(D) if

SOLUTION (A) Let

(B) Let

(C)

Let .

=

-312t + 121t2

+ t + 424

dudt

= 2t + 1 = -31t2+ t + 42-412t + 12

nun - 1 dudt

= -31t2+ t + 42-41t2

+ t + 42¿u = t2

+ t + 4, n = -3=

d

dt 1t2

+ t + 42-3

d

dt

1

1t2+ t + 423

= 21x21x3+ 426

dudx

= 3x2 = 71x3+ 426 3x2

nun - 1 dudx

y¿ = 71x3+ 4261x3

+ 42¿u = 1x3

+ 42, n = 7. y = 1x3+ 427

= 1213x + 123dudx

= 3 = 413x + 123 3nun - 1

dudx

f¿1x2 = 413x + 12313x + 12¿u = 3x + 1, n = 4. f1x2 = 13x + 124

h1w2 = 23 - wdh

dw

d

dt

1

1t2+ t + 423

y = 1x3+ 427y¿

f1x2 = 13x + 124f¿1x2


(D) Let

= -

1

213 - w21>2 or -

1

223 - w

dudw

= - 1 =

12

13 - w2-1>21-12

nun - 1 dudw

dh

dw =

12

13 - w2-1>213 - w2¿u = 3 - w, n =

12

h1w2 = 23 - w = 13 - w21>2244 C H A P T E R 4 Additional Derivative Topics

MATCHED PROBLEM 3 Find the indicated derivatives:

(A) if

(B) if

(C) (D) if g1w2 = 24 - wdg

dw

d

dt

1

1t2+ 422

y = 1x4- 525y¿

h1x2 = 15x + 223h¿1x2

Notice that we used two steps to differentiate each function in Example 3. First,we applied the general power rule; then we found du�dx. As you gain experiencewith the general power rule, you may want to combine these two steps. If you do this,be certain to multiply by du�dx. For example,

Correct

is missingdu>dx = 5x4 d

dx 1x5

+ 124 Z 41x5+ 123

d

dx 1x5

+ 124 = 41x5+ 1235x4

If we let then and the general power rule reduces to the (ordinary)power rule discussed in Section 3-5. Compare the following:

Yes—power rule

Yes—general power rule

Unless so that du>dx = 1u1x2 = x + k, d

dx un

Z nun - 1

d

dx un

= nun - 1 du

dx

d

dx xn

= nxn - 1

du>dx = 1,u1x2 = x,

I N S I G H T

� The Chain RuleWe have used the general power rule to find derivatives of composite functions of theform , where is a power function. But what if f is not a power func-tion? Then a more general rule, the chain rule, enables us to compute the derivativesof many composite functions of the form

Suppose that

is a composite of f and g, where

y = f1u2 and u = g1x2

y = m1x2 = f3g1x24f(g(x)).

f(u) = unf(g(x))


We would like to express the derivative in terms of the derivatives of f and g.From the definition of a derivative (see Section 3-4), we have

(4) = limh:0

cf3g1x + h24 - f3g1x24g1x + h2 - g1x2 #

g1x + h2 - g1x2h

d

= limh:0

cf3g1x + h24 - f3g1x24h

#

g1x + h2 - g1x2g1x + h2 - g1x2 d

= limh:0

f3g1x + h24 - f3g1x24h

dy

dx= lim

h:0

m1x + h2 - m1x2h

dy>dx


Substitute and

Multiply by 1 =

g1x + h2 - g1x2g1x + h2 - g1x2.

m1x2 = f3g1x24.m1x + h2 = f3g1x + h24

We recognize the second factor in equation (4) as the difference quotient for g(x). To interpret the first factor as the difference quotient for f(u), we let

Since we can write

Substituting in equation (1), we now have

(5)

If we assume that as we can find the limit of eachdifference quotient in equation (5):

This result is correct under rather general conditions and is called the chain rule, butour “derivation” is superficial, because it ignores a number of hidden problems. Sincea formal proof of the chain rule is beyond the scope of this book, we simply state itas follows:

=

dy

du du

dx

= f¿1u2g¿1x2 dy

dx= c lim

k:0

f1u + k2 - f1u2k

d c limh:0

g1x + h2 - g1x2h

d

h : 0,k = 3g1x + h2 - g1x24: 0

dy

dx= lim

h:0 cf1u + k2 - f1u2

k#g1x + h2 - g1x2

hd

u + k = g1x2 + g1x + h2 - g1x2 = g1x + h2u = g1x2,k = g1x + h2 - g1x2.

THEOREM 2 CHAIN RULEIf and define the composite function

then

or, equivalently,

m¿1x2 = f¿3g1x24g¿1x2 provided that f¿3g1x24 and g¿1x2 exist

dy

dx=

dy

du du

dx provided that

dy

du and

du

dx exist

y = m1x2 = f3g1x24u = g1x2y = f1u2

E X A M P L E 4 Using the Chain Rule Find dy�du, du�dx, and dy�dx (express dy�dx as a functionof x) for

(A) and

(B) and

(C) and u = x2- 4x + 2y = ln u

u = 2x3+ 5y = eu

u = 3x2+ 1y = u3>2


SOLUTION (A)

=

32

u1>216x2 = 9x13x2+ 121>2

dy

dx=

dy

du du

dx

dy

du=

32

u1>2 and du

dx= 6x


Basic derivative rules

Chain rule

Since u = 3x2+ 1

(B)

= eu16x22 = 6x2e2x3+ 5

dy

dx=

dy

du du

dx

dy

du= eu and

du

dx= 6x2 Basic derivative rules

Chain rule

Since u = 2x3+ 5

(C)

=

1u

12x - 42 =

2x - 4

x2- 4x + 2

dy

dx=

dy

du du

dx

dy

du=

1u

and du

dx= 2x - 4 Basic derivative rules

Chain rule

Since u = x2- 4x + 2

MATCHED PROBLEM 4 Find dy�du, du�dx, and dy�dx (express dy�dx as a function of x) for

(A) and

(B) and

(C) and u = x2+ 9x + 4y = ln u

u = 3x4+ 6y = eu

u = 2x3+ 4y = u-5

Explore & Discuss 2 Let Use the chain rule and the graphs in Figures 1 and 2 tofind

(A) f(4) (B) g(6) (C) m(6)

(D) (4) (E) (6) (F) (6)m¿g¿f¿

m1x2 = f3g1x24.

u105

y

30

20

10

y � f (u)

y � 2u � 4

(4, 12)

FIGURE 1

u � g(x)

u � 0.4x � 1.6

(6, 4)

x105

u

6

4

2

FIGURE 2

The chain rule can be extended to compositions of three or more functions. Forexample, if and then

dy

dx=

dy

dw dwdu

dudx

u = h1x2,y = f1w2, w = g1u2,



E X A M P L E 5 Using the Chain Rule For find dy�dx.

SOLUTION Note that h is of the form where and Thus,

Since

Since

=

2x

1ln x2e1 +1ln x22

u = ln x = e1 +1ln x2212 ln x2a 1xb

w = 1 + u2 = e1 + u212u2a 1xb

= ew12u2a 1xb

dy

dx =

dy

dw dw

du du

dx

u = ln x.w = 1 + u2y = ew,

y = h1x2 = e1 +1ln x22,

MATCHED PROBLEM 5 For find dy�dx.y = h1x2 = 3ln11 + ex243,

The chain rule generalizes basic derivative rules. We list three general derivativerules here for convenient reference [the first, equation (6), is the general power ruleof Theorem 1].

General Derivative Rules

(6)

(7)

(8)

Unless directed otherwise, you now have a choice between the chain rule and thegeneral derivative rules. However, practicing with the chain rule will help prepareyou for concepts that appear later in the text. Examples 4 and 5 illustrate the chainrule method, and the next example illustrates the general derivative rules method.

d

dx ef1x2

= ef1x2f¿1x2

d

dx ln3f1x24 =

1f1x2 f¿1x2

d

dx 3f1x24n = n3f1x24n - 1f¿1x2

E X A M P L E 6 Using General Derivative Rules

(A) Using equation (5)

(B) Using equation (4)

(C) Using equation (3)

Using equation (5)

= 6xex211 + ex222 = 311 + ex222ex212x2 = 311 + ex222ex2

d

dx x2

d

dx 11 + ex223 = 311 + ex222

d

dx 11 + ex22

=

1

x2+ 9

2x =

2x

x2+ 9

d

dx ln1x2

+ 92 =

1

x2+ 9

d

dx 1x2

+ 92 = e2x122 = 2e2x

d

dx e2x

= e2x d

dx 2x



2. (A) (B)[Note: There are other correct answers.]

3. (A)

(B)

(C)

(D)

4. (A)

(B)

(C)

5.

6. (A) (B) (C) - 8xe-x2

(2 + e-x2

)36xe3x2+ 23x2

+ 2x3

+ 2x

3ex[ln(1 + ex)]2

1 + ex

dy

du=

1u

, dudx

= 2x + 9, dy

dx=

2x + 9x2

+ 9x + 4

dy

du= eu,

dudx

= 12x3, dy

dx= 12x3e3x4

+ 6

dy

dx= - 5u-4,

dudx

= 6x2, dy

dx= - 30x2(2x3

+ 4)-6

-1>(224 - w)

-4t>(t2+ 4)3

20x3(x4- 5)4

15(5x + 2)2

f1u2 = 23 u, u = 1 + x3f1u2 = 50eu, u = -2x

f3g1x24 = 2ex, g3f1u24 = e2u


MATCHED PROBLEM 6 Find

(A) (B) (C)d

dx 12 + e-x224d

dx e3x2

+ 2d

dx ln1x3

+ 2x2

For many of the problems in this exercise set, the answers inthe back of the book include both an unsimplified form and asimplified form. When checking your work, first check that youapplied the rules correctly, and then check that you performedthe algebraic simplification correctly.

A In Problems 1–4, find f[g(x)].

1.

2.

3.

4.

In Problems 5–8, write each composite function in the formand

5.

6.

7.

8.

In Problems 9–16, replace the ? with an expression that willmake the indicated equation valid.

9.

10.

11.

12.d

dx 13x2

+ 725 = 513x2+ 724 ?

d

dx 14 - 2x223 = 314 - 2x222 ?

d

dx 15 - 2x26 = 615 - 2x25 ?

d

dx 13x + 424 = 413x + 423 ?

y = ex4+ 2x2

+ 5

y = e1 + x + x2

y = 12x3+ x + 325

y = 13x2- x + 524u = g1x2.y = f1u2

g1u2 = eu; g1x2 = 3x3

f1u2 = eu; g1x2 = -x2

f1u2 = u4; g1x2 = 1 - 4x3

f1u2 = u3; g1x2 = 3x2+ 2

13.

14.

15.

16.

In Problems 17–36, find and simplify.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

In Problems 37–42, find and the equation of the linetangent to the graph of f at the indicated value of x. Find thevalue(s) of x where the tangent line is horizontal.

37.

38. f1x2 = 13x - 124; x = 1

f1x2 = 12x - 123; x = 1

f ¿(x)

f1x2 = (x - 2 ln x)4f1x2 = (1 + ln x)3

f1x2 = 2 ln(x2- 3x + 4)f1x2 = 3 ln(1 + x2)

f1x2 = (x5+ 2)-3f1x2 = (x4

+ 1)-2

f1x2 = (4x + 3)1>2f1x2 = (2x - 5)1>2f1x2 = ex2

+ 3x + 1f1x2 = 3e-6x

f1x2 = 6e-2xf1x2 = e5x

f1x2 = (5x2- 3)6f1x2 = (3x2

+ 5)5

f1x2 = (6 - 0.5x)4f1x2 = (4 + 0.2x)5

f1x2 = (9 - 5x)2f1x2 = (5 - 2x)4

f1x2 = (3x - 7)5f1x2 = (2x + 5)3

f¿1x2

d

dx ln1x - x32 =

1

x - x3 ?

d

dx ln(x4

+ 1) =

1

x4+ 1

?

d

dx e4x-2

= e4x-2 ?

d

dx ex2

+ 1= ex2

+ 1 ?

Exercise 4-4



39.

40.

41.

42.

B In Problems 43–62, find the indicated derivative and simplify.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

In Problems 63–68, find and find the equation of the line tangent to the graph of f at the indicated value of x.

63.

64.

65.

66.

67.

68. f1x2 = e1x; x = 1

f1x2 = 2ln x; x = e

f1x2 =

x4

13x - 822; x = 4

f1x2 =

x

12x - 523; x = 3

f1x2 = x211 - x24; x = 2

f1x2 = x14 - x23; x = 2

f¿1x2g¿1t2 if g1t2 =

323 t - t2

f¿1t2 if f1t2 =

42t2- 3t

dy

dx if y = a 1

x2 - 5b-2

dy

dx if y = 132x - 125

d

dw

1

1w2- 226

d

dw

1

1w3+ 425

y¿ if y = [ln(x2+ 3)]3>2

y¿ if y = ln(x2+ 3)3>2

G¿1t2 if G1t2 = (1 - e2t)2

F¿1t2 if F1t2 = (et2+ 1)3

d

dx [x4 ln(1 + x4)]

d

dx ln(1 + x)

x3

h¿1x2 if h1x2 =

e2x

x2+ 9

g¿1x2 if g1x2 = 4xe3x

dg

dw if g1w2 = 23 3w - 7

dh

dw if h1w2 = 2w2

+ 8

d

dt 31t3

+ t22-2

d

dt 21t2

+ 3t2-3

y¿ if y = 21x3+ 625

y¿ if y = 31x2- 224

f1x2 = ln11 - x2+ 2x42; x = 1

f1x2 = 5ex2- 4x + 1; x = 0

f1x2 = 12x + 821>2; x = 4

f1x2 = 14x - 321>2; x = 3 In Problems 69–74, find and find the value(s) of x wherethe tangent line is horizontal.

69. 70.

71. 72.

73. 74.

75. Suppose a student reasons that the functionsand must

have the same derivative, since he has entered f(x), g(x),and into a graphing calculator, but only three

graphs appear (see the figure). Is his reasoning correct?Are and the same function? Explain.g¿1x2f¿1x2

g¿1x2f¿1x2,g1x2 = 4 ln1x2

+ 32f1x2 = ln351x2+ 3244

f1x2 = 2x2+ 4x + 5f1x2 = 2x2

- 8x + 20

f1x2 =

x - 1

1x - 323f1x2 =

x

12x + 522f1x2 = x31x - 724f1x2 = x21x - 523

f¿1x2

(A)

Figure for 75

�1

(B)

0

20

10

(A)

Figure for 76

1

(B)

1

2

4

76. Suppose a student reasons that the functionsand

must have the same derivative, since she has entered f(x),g(x), and into a graphing calculator, but onlythree graphs appear (see the figure). Is her reasoningcorrect? Are and the same function? Explain.g¿1x2f¿1x2

g¿1x2f¿1x2,g1x2 = 1x + 121>3f1x2 = 1x + 12 ln1x + 12 - x

C In Problems 77–92, find each derivative and simplify.

77. 78.

79. 80.

81. 82.

83. 84.

85. 86.

87. 88.

89. 90.

91. 92.ddxB 4x + 1

2x2+ 1

ddx2(2x - 1)3(x2

+ 3)4

ddx

x22x2

+ 1

ddx

2x2x - 3

ddx

10ln xddx

2x3- x2

+ 4x + 1

ddx

log5(5x2- 1)

ddx

log3(4x3+ 5x + 7)

ddx

81 - 2x2ddx

10x2+ x

ddx

log(x3- 1)

ddx

log2(3x2- 1)

d

dx

3x2

1x2+ 523

d

dx

1x3- 724

2x3

d

dx 32x21x3

- 3244d

dx 33x1x2

+ 1234



93. Cost function. The total cost (in hundreds of dollars) ofproducing x calculators per day is

(see the figure).

C1x2 = 10 + 22x + 16 0 … x … 50

97. Drug concentration. The concentration of a drug in thebloodstream t hours after injection is given approxi-mately by

where C(t) is concentration in milligrams per milliliter.

(A) What is the rate of change of concentration after1 hour? After 4 hours?

(B) Graph C.

98. Water pollution. The use of iodine crystals is a popularway of making small quantities of nonpotable water safeto drink. Crystals placed in a 1-ounce bottle of water willdissolve until the solution is saturated. After saturation,half of the solution is poured into a quart container ofnonpotable water, and after about an hour, the water isusually safe to drink. The half-empty 1-ounce bottle isthen refilled, to be used again in the same way. Supposethat the concentration of iodine in the 1-ounce bottle tminutes after the crystals are introduced can be approxi-mated by

where C(t) is the concentration of iodine in microgramsper milliliter.

(A) What is the rate of change of the concentration after1 minute? After 4 minutes?

(B) Graph C for

99. Blood pressure and age. A research group using hospi-tal records developed the following approximate mathe-matical model relating systolic blood pressure and age:

Here, P(x) is pressure, measured in millimeters ofmercury, and x is age in years. What is the rate ofchange of pressure at the end of 10 years? At the endof 30 years? At the end of 60 years?

100. Biology. A yeast culture at room temperature (68°F)is placed in a refrigerator maintaining a constanttemperature of 38°F. After t hours, the temperature Tof the culture is given approximately by

What is the rate of change of temperature of the cultureat the end of 1 hour? At the end of 4 hours?

101. Learning. In 1930, L. L. Thurstone developed thefollowing formula to indicate how learning time Tdepends on the length of a list n:

Here, a, c, and k are empirical constants. Suppose that,for a particular person, the time T (in minutes) requiredto learn a list of length n is

(A) Find dT�dn.

(B) Find and and interpret the results.f¿1272,f¿1112

T = f1n2 = 2n2n - 2

T = f1n2 =

c

k n2n - a

T = 30e-0.58t+ 38 t Ú 0

P1x2 = 40 + 25 ln1x + 12 0 … x … 65

0 … t … 5.

C1t2 = 25011 - e-t2 t Ú 0

C1t2 = 4.35e-t 0 … t … 5

C(x)

x10 20 30 40 50

Production

Cos

t (hu

ndre

d do

llars

)

5

10

15

20

25

Figure for 93

(A) Find .(B) Find and and interpret the results.

94. Cost function. The total cost (in hundreds of dollars) ofproducing x cameras per week is

(A) Find .(B) Find and and interpret the results.

95. Price–supply equation. The number x of stereo speakersa retail chain is willing to sell per week at a price of $p isgiven by

(see the figure).

(A) Find dx�dp.

(B) Find the supply and the instantaneous rate of changeof supply with respect to price when the price is $75.Write a brief verbal interpretation of these results.

x = 802p + 25 - 400 20 … p … 100

C¿1242,C¿1152C¿1x2

C1x2 = 6 + 24x + 4 0 … x … 30

C¿1422,C¿1242C¿1x2

x

p

Supp

ly/D

eman

d

200

400

600

800

20 40 60 80 100Price (dollars)

x � 1,000 � 60�p � 25

x � 80�p � 25 � 400

Figure for 95 and 96

Applications

96. Price–demand equation. The number x of stereo speak-ers people are willing to buy per week from a retail chainat a price of $p is given by

(see the previous figure).

(A) Find dx�dp.

(B) Find the demand and the instantaneous rate of changeof demand with respect to price when the price is $75.Write a brief verbal interpretation of these results.

x = 1,000 - 602p + 25 20 … p … 100


Section 4-5 IMPLICIT DIFFERENTIATION� Special Function Notation� Implicit Differentiation

� Special Function NotationThe equation

(1)

defines a function f with y as a dependent variable and x as an independent variable.Using function notation, we would write

In order to reduce to a minimum the number of symbols involved in a discussion, wewill often write equation (1) in the form

where y is both a dependent variable and a function symbol. This is a convenientnotation, and no harm is done as long as one is aware of the double role of y. Otherexamples are

This type of notation will simplify much of the discussion and work that follows.Until now, we have considered functions involving only one independent variable.

There is no reason to stop there: The concept can be generalized to functions involvingtwo or more independent variables, and this will be done in detail in Chapter 8. Fornow, we will “borrow” the notation for a function involving two independentvariables. For example,

specifies a function F involving two independent variables.

� Implicit DifferentiationConsider the equation

(2)

and the equation obtained by solving equation (2) for y in terms of x,

(3)

Both equations define the same function with x as the independent variable and y asthe dependent variable. On the one hand, we can write, for equation (3),

where

(4)

and we have an explicit (directly stated) rule that enables us to determine y for eachvalue of x. On the other hand, the y in equation (2) is the same y as in equation (3),and equation (2) implicitly gives (implies, though does not directly express) y as afunction of x. Thus, we say that equations (3) and (4) define the function f explicitlyand equation (2) defines f implicitly.

The direct use of an equation that defines a function implicitly to find the deriva-tive of the dependent variable with respect to the independent variable is called

f1x2 = 2 - 3x2

y = f1x2

y = 2 - 3x2

3x2+ y - 2 = 0

F1x, y2 = x2- 2xy + 3y2

- 5

r =

1

1s2- 3s22>3 = r1s2

z = 2u2- 3u = z1u2

x = 2t2- 3t + 1 = x1t2

y = 2 - 3x2= y1x2

y = f1x2 or f1x2 = 2 - 3x2

y = 2 - 3x2

S e c t i o n 4 - 5 Implicit Differentiation 251


implicit differentiation. Let us differentiate equation (2) implicitly and equation (3)directly, and compare results.

Starting with

we think of y as a function of x—that is, —and write

Then we differentiate both sides with respect to x:

6x + y¿ - 0 = 0

d

dx 3x2

+

d

dx y1x2 -

d

dx 2 = 0

d

dx 313x2

+ y1x2 - 224 =

d

dx 0

3x2+ y1x2 - 2 = 0

y = y1x23x2

+ y - 2 = 0


Since y is a function of x, but is notexplicitly given, we simply write

to indicate its derivative.ddx

y1x2 = y¿

Now we solve for

Note that we get the same result if we start with equation (3) and differentiate directly:

Why are we interested in implicit differentiation? In general, why do we not solvefor y in terms of x and differentiate directly? The answer is that there are many equa-tions of the form

(5)

that are either difficult or impossible to solve for y explicitly in terms of x (try it foror for for example). But it can be shown that,

under fairly general conditions on F, equation (5) will define one or more functionsin which y is a dependent variable and x is an independent variable. To find underthese conditions, we differentiate equation (5) implicitly.

Explore & Discuss 1 (A) How many tangent lines are there to the graph in Figure 1 when When When When When x = 6?x = 4?x = 2?x = 1?

x = 0?

y¿

ey- y = 3x,x2y5

- 3xy + 5 = 0

F1x, y2 = 0

y¿ = -6x

y = 2 - 3x2

y¿ = -6x

y¿:

x5

y

�5

5

FIGURE 1

(B) Sketch the tangent lines referred to in part (A), and estimate each of theirslopes.

(C) Explain why the graph in Figure 1 is not the graph of a function.



E X A M P L E 1 Differentiating Implicitly Given

(6)

find and the slope of the graph at

SOLUTION We start with the graph of (a circle, as shown in Fig. 2) so that wecan interpret our results geometrically. From the graph, it is clear that equation (6)does not define a function. But with a suitable restriction on the variables, equation (6)can define two or more functions. For example, the upper half and the lower half ofthe circle each define a function. On each half-circle, a point that corresponds to

is found by substituting into equation (6) and solving for y:

Thus, the point (3, 4) is on the upper half-circle, and the point is on the lowerhalf-circle. We will use these results in a moment. We now differentiate equation (6)implicitly, treating y as a function of x [i.e.,

Use the chain rule.

Solve for in terms of x and y.

Leave the answer in terms of x and y.

We have found without first solving for y in terms of x. And byleaving in terms of x and y, we can use to find for any point on thegraph of (except where ). In particular, for we foundthat (3, 4) and are on the graph; thus, the slope of the graph at (3, 4) is

The slope of the graph at (3, 4)

and the slope at is

The slope of the graph at

The symbol

is used to indicate that we are evaluating at and The results are interpreted geometrically in Figure 3 on the original graph.

In Example 1, the fact that is given in terms of both x and y is not a greatdisadvantage. We have only to make certain that when we want to evaluate for aparticular value of x and y, say, the ordered pair must satisfy the originalequation.

1x0, y02,yœ

y¿

y = b.x = ay¿

y¿ ƒ 1a, b2

13, -42y¿ ƒ 13, -42 = -3

- 4 =34

13, -42y¿ ƒ 13, 42 = -

34

13, -42 x = 3,y = 0x2+ y2

- 25 = 0y¿y¿ = -x>yy¿

x2+ y2

- 25 = 0y¿

y¿ = -

xy

y¿ = -

2x

2y

y ¿ 2x + 2yy¿ = 0

2x + 23y1x242 - 1y¿1x2 - 0 = 0

d

dx x2

+

d

dx 3y1x242 -

d

dx 25 = 0

d

dx5x2

+ 3y1x242 - 256 =

d

dx 0

x2+ 3y1x242 - 25 = 0

x2+ y2

- 25 = 0

y = y1x2]:13, -42

y = ;4

y2= 16

1322 + y2= 25

x2+ y2

- 25 = 0

x = 3x = 3

x2+ y2

- 25 = 0

x = 3.y¿

F1x, y2 = x2+ y2

- 25 = 0

x2 � y2 � 25

x

y

�4

40�4

4

FIGURE 2

x2 � y2 � 25

x

y

�4

40�4

4 (3, 4)

(3, �4)xyy� � �

Slope � �!

Slope � !

FIGURE 3


Now find the slope at each point:

Equation of tangent line at Equation of tangent line at (1, 2):

y = -23 x +

83 y = -

13 x -

23

y - 2 = -23 x +

23 y + 1 = -

13 x +

13

y - 2 = -231x - 12 y + 1 = -

131x - 12

y - y1 = m1x - x12 y - y1 = m1x - x1211, -12:

y¿ƒ 11,22 =

1222 - 21121 - 2112122 =

4 - 21 - 4

=

2-3

= -

23

y¿ƒ 11, -12 =

1-122 - 21121 - 21121-12 =

1 - 21 + 2

=

-13

= -

13


In the solution for Example 1, notice that the derivative of with respect to x is notjust 2y. When first learning about implicit differentiation, it is a good idea to replace ywith y(x) in the original equation to emphasize that y is a function of x.

2yy¿,y2

I N S I G H T

MATCHED PROBLEM 1 Graph find by implicit differentiation, and find the slope of thegraph when x = 5.

y¿x2+ y2

- 169 = 0,

E X A M P L E 2 Differentiating Implicitly Find the equation(s) of the tangent line(s) to thegraph of

(7)

at the point(s) where

SOLUTION We first find y when

Thus, there are two points on the graph of (7) where namely, and(1, 2). We next find the slope of the graph at these two points by differentiatingequation (7) implicitly:

y¿ =

y2- 2x

1 - 2xy

11 - 2xy2y¿ = y2- 2x

y¿ - 2xyy¿ = y2- 2x

y¿ - 2xyy¿ - y2+ 2x = 0

y¿ - 1x # 2yy¿ + y22 + 2x = 0

d

dx y -

d

dx xy2

+

d

dx x2

+

d

dx 1 =

d

dx 0

y - xy2+ x2

+ 1 = 0

11, -12x = 1,

y = -1, 2

1y - 221y + 12 = 0

y2- y - 2 = 0

y - y2+ 2 = 0

y - 112y2+ 1122 + 1 = 0

y - xy2+ x2

+ 1 = 0

x = 1:

x = 1.

y - xy2+ x2

+ 1 = 0

Use the product rule and the

chain rule for

Solve for by getting allterms involving on one side.y ¿

y ¿

ddx

xy2.


Explore & Discuss 2 The slopes of the tangent lines to when can be foundeither (1) by differentiating the equation implicitly or (2) by solving for y explicitlyin terms of x (using the quadratic formula) and then computing the derivative.Which of the two methods is more efficient? Explain.

x = 0y2+ 3xy + 4x = 9


MATCHED PROBLEM 2 Repeat Example 2 for at x = 1.x2+ y2

- xy - 7 = 0

E X A M P L E 3 Differentiating Implicitly Find for defined implicitly by

and evaluate at

SOLUTION It is important to remember that x is the dependent variable and t is the indepen-dent variable. Therefore, we differentiate both sides of the equation with respect tot (using product and chain rules where appropriate) and then solve for

Differentiate implicitly with respect to t.

Clear fractions.

Solve for Factor out

Now we evaluate at as requested:

=

1-1

= -1

x¿ƒ 10,12 =

1122e0- 1 ln 1

0 - 1e0

1t, x2 = 10, 12,x¿

x¿ =

x2et- x ln x

t - xet

1t - xet2x¿ = x2et- x ln x

x¿. tx¿ - xe¿x¿ = x2et- x ln x

x¿. tx¿ + x ln x = x2et+ xetx¿

x Z 0x # t x¿

x+ x # ln x = x # xet

+ x # etx¿

t x¿

x+ ln x = xet

+ x¿et

d

dt 1t ln x2 =

d

dt 1xet2 -

d

dt 1

t ln x = xet- 1

x¿:

1t, x2 = 10, 12.x¿

t ln x = xet- 1

x = x1t2x¿

MATCHED PROBLEM 3 Find for defined implicitly by

and evaluate at 1t, x2 = 11, 02.x¿

1 + x ln t = tex

x = x1t2x¿

Answers to Matched Problems 1. . When thus, and

2. 3. x¿ =

tex- x

t ln t - t2ex; x¿ ƒ 11,02 = -1y¿ =

y - 2x

2y - x; y =

45 x -

145 , y =

15 x +

145

y¿ ƒ 15,-122 =5

12y¿ ƒ 15,122 = -512x = 5, y = ;12;y¿ = -x>y



A In Problems 1–4, find in two ways:

(A) Differentiate the given equation implicitly and then solvefor

(B) Solve the given equation for y and then differentiate directly.

1.

2.

3.

4.

In Problems 5–22, use implicit differentiation to find and evaluate at the indicated point.

5.

6.

7.

8.

9.

10.

B11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

In Problems 23 and 24, find for defined implicitlyby the given equation. Evaluate at the indicated point.

23.

24.

Problems 25 and 26 refer to the equation and graph shown inthe figure.

x3- tx2

- 4 = 0; 1-3, -22x2

- t2x + t3+ 11 = 0; 1-2, 12

x¿

x = x1t2x¿

xey- y = x2

- 2; 12, 02x ln y + 2y = 2x3; 11, 12ln y = 2y2

- x; 12, 12x3

- y = ln y; 11, 12x2

- y = 4ey; 12, 02ey

= x2+ y2; 11, 02

2x3y - x3+ 5 = 0; 1-1, 32

x2y - 3x2- 4 = 0; 12, 42

2y + xy - 1 = 0; 1-1, 122xy + y + 2 = 0; 1-1, 223xy - 2x - 2 = 0; 12, 12xy - 6 = 0; 12, 32

y2- y - 4x = 0; 10, 12

y2+ 2y + 3x = 0; 1-1, 12

y2+ x3

+ 4 = 0; 1-2, 22x2

- y3- 3 = 0; 12, 12

5x3- y - 1 = 0; 11, 42

y - 5x2+ 3 = 0; 11, 22

y¿

y¿

2x3+ 5y - 2 = 0

3x2- 4y - 18 = 0

-2x + 6y - 4 = 0

3x + 5y + 9 = 0

y¿.

y¿

25. Use implicit differentiation to find the slopes of thetangent lines at the points on the graph where Check your answers by visually estimating the slopeson the graph in the figure.

26. Find the slopes of the tangent lines at the points on thegraph where Check your answers by visuallyestimating the slopes on the graph in the figure.

In Problems 27–30, find the equation(s) of the tangent line(s) tothe graphs of the indicated equations at the point(s) with thegiven value of x.

27.

28.

29.

30.

31. If find in two ways, first by differentiatingimplicitly and then by solving for y explicitly in termsof x. Which method do you prefer? Explain.

32. Explain the difficulty that arises in solvingfor y as an explicit function of x.

Find the slope of the tangent line to the graph of theequation at the point (0, 1).

C In Problems 33–40, find and the slope of the tangent line tothe graph of each equation at the indicated point.

33.

34.

35.

36.

37.

38.

39.

40.

41. Find the equation(s) of the tangent line(s) at the point(s)on the graph of the equation

where Round all approximate values to twodecimal places.

42. Refer to the equation in Problem 41. Find the equation(s)of the tangent line(s) at the point(s) on the graph where

Round all approximate values to two decimalplaces.y = -1.

x = 1.

y3- xy - x3

= 2

exy- 2x = y + 1; 10, 02

ln1xy2 = y2- 1; 11, 12

62y3+ 1 - 2x3>2

- 2 = 0; 14, 2227 + y2

- x3+ 4 = 0; 12, 32

12x - y24 - y3= 8; 1-1, -22

1x - 2y23 = 2y2- 3; 11, 12

1y - 324 - x = y; 1-3, 4211 + y23 + y = x + 7; 12, 12

y¿

x3+ y + xey

= 1

y¿xey= 1,

xy2- y - 2 = 0; x = 1

y2- xy - 6 = 0; x = 1

3x + xy + 1 = 0; x = -1

xy - x - 4 = 0; x = 2

x = 0.2.

x = 1.6.

Exercise 4-5

x

y

21

2

(x � 1)2 � (y � 1)2 � 1

1


S e c t i o n 4 - 6 Related Rates 257

For the demand equations in Problems 43–46, find the rate of change of p with respect to x by differentiating implicitly(x is the number of items that can be sold at a price of $p).

43.

44.

45.

46. x = 23 1,500 - p3

x = 210,000 - p2

x = p3- 3p2

+ 200

x = p2- 2p + 1,000

47. Biophysics. In biophysics, the equation

is called the fundamental equation of muscle contraction,where m, n, and k are constants and V is the velocity ofthe shortening of muscle fibers for a muscle subjected toa load L. Find dL�dV by implicit differentiation.

48. Biophysics. In Problem 47, find dV�dL by implicitdifferentiation.

1L + m21V + n2 = k

Applications

Section 4-6 RELATED RATESThe workers in a union are concerned that the rate at which wages are increasing islagging behind the rate of increase in the company’s profits. An automobile dealerwants to predict how badly an anticipated increase in interest rates will decrease hisrate of sales. An investor is studying the connection between the rate of increase inthe Dow Jones average and the rate of increase in the gross domestic product overthe past 50 years.

In each of these situations, there are two quantities—wages and profits in the firstinstance, for example—that are changing with respect to time. We would like todiscover the precise relationship between the rates of increase (or decrease) of thetwo quantities. We will begin our discussion of such related rates by consideringsome familiar situations in which the two quantities are distances and the two ratesare velocities.

26 ft

x

y

E X A M P L E 1 Related Rates and Motion A 26-foot ladder is placed against a wall (Fig. 1).If the top of the ladder is sliding down the wall at 2 feet per second, at what rate isthe bottom of the ladder moving away from the wall when the bottom of the ladderis 10 feet away from the wall?

SOLUTION Many people reason that since the ladder is of constant length, the bottom of theladder will move away from the wall at the rate that the top of the ladder is movingdown the wall. This is not the case, as we will see.

At any moment in time, let x be the distance of the bottom of the ladder from thewall and let y be the distance of the top of the ladder from the ground (see Fig. 1).Both x and y are changing with respect to time and can be thought of as functions oftime; that is, and Furthermore, x and y are related by thePythagorean relationship:

(1)

Differentiating equation (1) implicitly with respect to time t and using the chain rulewhere appropriate, we obtain

(2)

The rates dx�dt and dy�dt are related by equation (2); hence, this type of problem isreferred to as a related-rates problem.

Now, our problem is to find dx�dt when feet, given that (y is decreasing at a constant rate of 2 feet per second). We have all the quantities weneed in equation (2) to solve for dx�dt, except y. When y can be found fromequation (1):

y = 2262- 102

= 24 feet

102+ y2

= 262

x = 10,

dy>dt = -2x = 10

2x dx

dt+ 2y

dy

dt= 0

x2+ y2

= 262

y = y1t2.x = x1t2

FIGURE 1


Substitute and into (2); then solve for dx�dt:

Thus, the bottom of the ladder is moving away from the wall at a rate of 4.8 feet persecond.

dx

dt=

-212421-2221102 = 4.8 feet per second

21102 dx

dt+ 212421-22 = 0

y = 24dy>dt = -2, x = 10,


In the solution to Example 1, we used equation (1) in two ways: first, to find an equationrelating dy�dt and dx�dt, and second, to find the value of y when These stepsmust be done in this order; substituting and then differentiating does not produceany useful results:

y¿ = 0

0 + 2yy¿ = 0

100 + y2= 262

x2+ y2

= 262

x = 10x = 10.

I N S I G H T

Substituting 10 for x has the effect of stopping the ladder.

The rate of change of a stationary object is always 0, but that is not the rate of change of the moving ladder.

MATCHED PROBLEM 1 Again, a 26-foot ladder is placed against a wall (Fig. 1). If the bottom of the ladder ismoving away from the wall at 3 feet per second, at what rate is the top moving downwhen the top of the ladder is 24 feet up the wall?

Explore & Discuss 1 (A) For which values of x and y in Example 1 is dx�dt equal to 2 (i.e., the samerate at which the ladder is sliding down the wall)?

(B) When is dx�dt greater than 2? Less than 2?

DEFINITION Suggestions for Solving Related-Rates ProblemsStep 1. Sketch a figure if helpful.

Step 2. Identify all relevant variables, including those whose rates are given andthose whose rates are to be found.

Step 3. Express all given rates and rates to be found as derivatives.

Step 4. Find an equation connecting the variables identified in step 2.

Step 5. Implicitly differentiate the equation found in step 4, using the chain rulewhere appropriate, and substitute in all given values.

Step 6. Solve for the derivative that will give the unknown rate.

E X A M P L E 2 Related Rates and Motion Suppose that two motorboats leave from the samepoint at the same time. If one travels north at 15 miles per hour and the other travelseast at 20 miles per hour, how fast will the distance between them be changing after2 hours?


SOLUTION First, draw a picture, as shown in Figure 2.All variables, x, y, and z, are changing with time. Hence, they can be thought of

as functions of time: and given implicitly. It now makessense to take derivatives of each variable with respect to time. From the Pythagoreantheorem,

(3)

We also know that

We would like to find dz�dt at the end of 2 hours—that is, when miles andmiles. To do this, we differentiate both sides of equation (3) with respect to

t and solve for dz�dt:

(4)

We have everything we need except z. From equation (3), when andwe find z to be 50. Substituting the known quantities into equation (4), we obtain

Thus, the boats will be separating at a rate of 25 miles per hour.

dz

dt= 25 miles per hour

21502 dz

dt= 214021202 + 213021152

y = 30,x = 40

2z dz

dt= 2x

dx

dt+ 2y

dy

dt

y = 30x = 40

dx

dt= 20 miles per hour and

dy

dt= 15 miles per hour

z2= x2

+ y2

z = z1t2,x = x1t2, y = y1t2,


MATCHED PROBLEM 2 Repeat Example 2 for the situation at the end of 3 hours.

E X A M P L E 3 Related Rates and Motion Suppose a point is moving along the graph of(Fig. 3). When the point is at its x coordinate is increasing at

the rate of 0.4 unit per second. How fast is the y coordinate changing at that moment?

SOLUTION Since both x and y are changing with respect to time, we can think of each as a func-tion of time, namely,

but restricted so that

(5)

Our problem now is to find dy�dt, given and Implicitlydifferentiating both sides of equation (5) with respect to t, we have

dy

dt= 0.3 unit per second

1-3210.42 + 4

dy

dt= 0

x dx

dt+ y

dy

dt= 0

2x dx

dt+ 2y

dy

dt= 0

x2+ y2

= 25

dx>dt = 0.4.x = -3, y = 4,

x2+ y2

= 25

x = x1t2 and y = y1t2

1-3, 42,x2+ y2

= 25

Divide both sides by 2.

Substitute andand solve for .dy>dtdx>dt = 0.4,

x = -3, y = 4,

Ex

yz

N

FIGURE 2

x2 � y2 � 25

x

y

�4

40�4

4(�3, 4)

FIGURE 3



MATCHED PROBLEM 3 A point is moving on the graph of When the point is at its y coor-dinate is decreasing by 2 units per second. How fast is the x coordinate changing atthat moment?

1-8, 42,y3= x2.

E X A M P L E 4 Related Rates and Business Suppose that for a company manufacturing tran-sistor radios, the cost, revenue, and profit equations are given by

Cost equationRevenue equationProfit equation

where the production output in 1 week is x radios. If production is increasing atthe rate of 500 radios per week when production is 2,000 radios, find the rate ofincrease in

(A) Cost (B) Revenue (C) Profit

SOLUTION If production x is a function of time (it must be, since it is changing with respect totime), then C, R, and P must also be functions of time. These functions are implicit-ly (rather than explicitly) given. Letting t represent time in weeks, we differentiateboth sides of each of the preceding three equations with respect to t and thensubstitute and to find the desired rates.

(A) Think: and

Differentiate both sides with respect to t.

Since when

Cost is increasing at a rate of $1,000 per week.

(B)

Since when

Revenue is increasing at a rate of $3,000 per week.

(C)

Results from parts (A) and (B)

Profit is increasing at a rate of $2,000 per week.

= $2,000 per week

= $3,000 - $1,000

dP

dt=

dR

dt-

dC

dt

P = R - C

dR

dt= 310 - 0.00212,0002415002 = $3,000 per week

x = 2,000,dx>dt = 500

dR

dt= 110 - 0.002x2

dx

dt

dR

dt= 10

dx

dt- 0.002x

dx

dt

dR

dt=

d

dt 110x2 -

d

dt 0.001x2

R = 10x - 0.001x2

dC

dt= 215002 = $1,000 per week

x = 2,000,dx>dt = 500

dC

dt= 0 + 2

dx

dt= 2

dx

dt

dC

dt=

d

dt 15,0002 +

d

dt 12x2

x = x1t2.C = C1t2 C = 5,000 + 2x

dx>dt = 500x = 2,000

P = R - C

R = 10x - 0.001x2

C = 5,000 + 2x


Explore & Discuss 2 (A) In Example 4, suppose that Find the time and produc-tion level at which the profit is maximized.

(B) Suppose that Find the time and production level atwhich the profit is maximized.

(C) Explain why it is unnecessary to know a formula for x(t) in order to deter-mine the production level at which the profit is maximized.


2.

3.

4. (A)

(B)

(C) dP>dt = -$2,000>wk

dR>dt = -$1,000>wk

dC>dt = $1,000>wk

dx>dt = 6 units>sec

dz>dt = 25 mi>hr

dy>dt = -1.25 ft>sec

x1t2 = t2+ 492t + 16.

x1t2 = 500t + 500.


MATCHED PROBLEM 4 Repeat Example 4 for a production level of 6,000 radios per week.

A In Problems 1–6, assume that and Find the indicated rate, given the other information.

1. when find dy�dt

2. when find dy�dt

3. when and find dx�dt

4. when and find dx�dt

5. when and find dy�dt

6. when andfind dx�dt

B7. A point is moving on the graph of When the

point is at (4, 9), its x coordinate is increasing by 4 unitsper second. How fast is the y coordinate changing at thatmoment?

8. A point is moving on the graph of When the point is at (3, 0), its y coordinate is decreasingby 2 units per second. How fast is its x coordinatechanging at that moment?

9. A boat is being pulled toward a dock as indicated in thefigure. If the rope is being pulled in at 3 feet per second,how fast is the distance between the dock and the boatdecreasing when it is 30 feet from the dock?

4x2+ 9y2

= 36.

xy = 36.

y = -1;x = 2x2

- 2xy - y2= 7; dy>dt = -1

y = 2;x = 1x2

+ 3xy + y2= 11; dx>dt = 2

y = -1.6;x = 1.2x2+ y2

= 4; dy>dt = 5

y = 0.8;x = -0.6x2

+ y2= 1; dy>dt = -4

x = 2;y = x3- 3; dx>dt = -2

x = 5;y = x2+ 2; dx>dt = 3

y = y1t2.x = x1t2 10. Refer to Problem 9. Suppose that the distance betweenthe boat and the dock is decreasing by 3.05 feet persecond. How fast is the rope being pulled in when theboat is 10 feet from the dock?

11. A rock thrown into a still pond causes a circular ripple. Ifthe radius of the ripple is increasing by 2 feet per second,how fast is the area changing when the radius is 10 feet?[Use ]

12. Refer to Problem 11. How fast is the circumference ofa circular ripple changing when the radius is 10 feet?[Use ]

13. The radius of a spherical balloon is increasing at therate of 3 centimeters per minute. How fast is thevolume changing when the radius is 10 centimeters?[Use ]

14. Refer to Problem 13. How fast is the surface area of thesphere increasing when the radius is 10 centimeters?[Use ]

15. Boyle’s law for enclosed gases states that if the volumeis kept constant, the pressure P and temperature T arerelated by the equation

where k is a constant. If the temperature is increasing at3 kelvins per hour, what is the rate of change of pressurewhen the temperature is 250 kelvins and the pressure is500 pounds per square inch?

16. Boyle’s law for enclosed gases states that if the tempera-ture is kept constant, the pressure P and volume V of agas are related by the equation

where k is a constant. If the volume is decreasing by5 cubic inches per second, what is the rate of change ofpressure when the volume is 1,000 cubic inches and thepressure is 40 pounds per square inch?

VP = k

P

T= k

S = 4pR2, p L 3.14.

V =43 pR3, p L 3.14.

C = 2pR, p L 3.14.

A = pR2, p L 3.14.

Exercise 4-6

Rope4 ft

Figure for 9 and 10



17. A 10-foot ladder is placed against a vertical wall. Supposethe bottom of the ladder slides away from the wall at aconstant rate of 3 feet per second. How fast is the top ofthe ladder sliding down the wall (negative rate) whenthe bottom is 6 feet from the wall? [Hint: Use thePythagorean theorem, where c is the lengthof the hypotenuse of a right triangle and a and b are thelengths of the two shorter sides.]

18. A weather balloon is rising vertically at the rate of 5 meters per second. An observer is standing on theground 300 meters from the point where the balloonwas released. At what rate is the distance between theobserver and the balloon changing when the balloon is400 meters high?

C19. A streetlight is on top of a 20-foot pole. A person who is

5 feet tall walks away from the pole at the rate of 5 feetper second. At what rate is the tip of the person’s shadowmoving away from the pole when he is 20 feet from thepole?

a2+ b2

= c2,

20. Refer to Problem 19. At what rate is the person’s shadowgrowing when he is 20 feet from the pole?

21. Helium is pumped into a spherical balloon at a constantrate of 4 cubic feet per second. How fast is the radiusincreasing after 1 minute? After 2 minutes? Is there anytime at which the radius is increasing at a rate of 100 feetper second? Explain.

22. A point is moving along the x axis at a constant rate of5 units per second. At which point is its distance from(0, 1) increasing at a rate of 2 units per second? At 4 unitsper second? At 5 units per second? At 10 units persecond? Explain.

23. A point is moving on the graph of insuch a way that its x coordinate is always increasing at arate of 3 units per second. How fast is the y coordinatechanging when the point crosses the x axis?

24. A point is moving on the graph of in such away that its y coordinate is always increasing at a rate of2 units per second. At which point(s) is the x coordinateincreasing at a rate of 1 unit per second?

x3+ y2

= 1

y = ex+ x + 1

25. Cost, revenue, and profit rates. Suppose that for acompany manufacturing calculators, the cost, revenue,and profit equations are given by

where the production output in 1 week is x calculators.If production is increasing at a rate of 500 calculators perweek when production output is 6,000 calculators, find therate of increase (decrease) in

(A) Cost (B) Revenue (C) Profit

26. Cost, revenue, and profit rates. Repeat Problem 25 for

where production is increasing at a rate of 500 calculatorsper week at a production level of 1,500 calculators.

27. Advertising. A retail store estimates that weekly sales sand weekly advertising costs x (both in dollars) arerelated by

The current weekly advertising costs are $2,000, andthese costs are increasing at the rate of $300 per week.Find the current rate of change of sales.

28. Advertising. Repeat Problem 27 for

29. Price–demand. The price p (in dollars) and demand x fora product are related by

2x2+ 5xp + 50p2

= 80,000

s = 50,000 - 20,000e-0.0004x

s = 60,000 - 40,000e-0.0005x

P = R - C

C = 72,000 + 60x R = 200x -

x2

30

P = R - C

C = 90,000 + 30x R = 300x -

x2

30

(A) If the price is increasing at a rate of $2 per month whenthe price is $30, find the rate of change of the demand.

(B) If the demand is decreasing at a rate of 6 units permonth when the demand is 150 units, find the rate ofchange of the price.

30. Price–demand. Repeat Problem 29 for

31. Pollution. An oil tanker aground on a reef is leakingoil that forms a circular oil slick about 0.1 foot thick (seethe figure). To estimate the rate dV�dt (in cubic feet perminute) at which the oil is leaking from the tanker, itwas found that the radius of the slick was increasing at0.32 foot per minute when the radius Rwas 500 feet. Find dV�dt, using p L 3.14.

(dR�dt = 0.32)

x2+ 2xp + 25p2

= 74,500

Applications

Oil slick

TankerR

A � pR2 V � 0.1 A

Figure for 31

32. Learning. A person who is new on an assembly lineperforms an operation in T minutes after x performancesof the operation, as given by

If operations per hour, where t is time inhours, find dT�dt after 36 performances of the operation.

dx�dt = 6

T = 6 a1 +

12xb


Section 4-7 ELASTICITY OF DEMAND� Relative Rate of Change� Elasticity of Demand

When will a price increase lead to an increase in revenue? To answer this questionand, more generally, to study relationships among price, demand, and revenue, econ-omists use the notion of elasticity of demand. In this section, we define the conceptsof relative rate of change and percentage rate of change, and we give an introductionto elasticity of demand.

� Relative Rate of ChangeExplore & Discuss 1 A broker is trying to sell you two stocks: Biotech and Comstat. The broker esti-

mates that Biotech’s earnings will increase $2 per year over the next several years,while Comstat’s earnings will increase only $1 per year. Is this sufficient infor-mation for you to choose between the two stocks? What other information mightyou request from the broker to help you decide?

Interpreting rates of change is a fundamental application of calculus. In Explore& Discuss 1, Biotech’s earnings are increasing at twice the rate of Comstat’s, but thatdoes not automatically make Biotech the better buy. The obvious information thatis missing is the cost of each stock. If Biotech costs $100 a share and Comstat costs$25 share, then which stock is the better buy? To answer this question, we introducetwo new concepts: relative rate of change and percentage rate of change.

S e c t i o n 4 - 7 Elasticity of Demand 263

DEFINITION Relative and Percentage Rates of Change

The relative rate of change of a function f(x) is

The percentage rate of change is 100 *

f¿1x2f1x2 .

f¿1x2f1x2 .

Because

the relative rate of change of f(x) is the derivative of the logarithm of f(x). This is alsoreferred to as the logarithmic derivative of f(x). Returning to Explore & Discuss 1,we can now write

Relative rate of change Percentage rate of change

Biotech or 2%

Comstat or 4%1

25= 0.04

2100

= 0.02

d

dx ln f1x2 =

f¿1x2f1x2

E X A M P L E 1 Relative Rate of Change Table 1 at the top of the next page lists the real GDP(gross domestic product expressed in billions of 1996 dollars) and population of theUnited States from 1995 to 2002. A model for the GDP is

where t is years since 1990. Find and graph the percentage rate of change of f(t) for5 … t … 12.

f1t2 = 300t + 6,000


SOLUTION If p(t) is the percentage rate of change of f(t), then

The graph of p(t) is shown in Figure 1 (graphing details omitted). Notice that p(t) isdecreasing, even though the GDP is increasing.

=

100t + 20

=

30,000300t + 6,000

p1t2 = 100 *

d

dx ln1300t + 6,0002


Real GDP Population Year (billions of 1996 dollars) (in millions)

1995 $7,540 262.765

1996 $7,810 265.19

1997 $8,150 267.744

1998 $8,500 270.299

1999 $8,850 272.82

2000 $9,190 275.306

2001 $9,210 277.803

2002 $9,440 280.306

TABLE 1

MATCHED PROBLEM 1 A model for the population data in Table 1 is

where t is years since 1990. Find and graph p(t), the percentage rate of change of f(t)for 5 … t … 12.

f1t2 = 2.5t + 250

� Elasticity of DemandLogarithmic derivatives and relative rates of change are used by economists to studythe relationship among price changes, demand, and revenue. Suppose the price $p andthe demand x for a certain product are related by the price–demand equation

(1)

In problems involving revenue, cost, and profit, it is customary to use the demandequation to express price as a function of demand. Since we are now interested in theeffects that changes in price have on demand, it will be more convenient to expressdemand as a function of price. Solving (1) for x, we have

Demand as a function of price = 500120 - p2 x = 10,000 - 500p

x + 500p = 10,000

t125

p(t)

4

2

FIGURE 1


or

(2)

Since x and p both represent nonnegative quantities, we must restrict p so thatFor most products, demand is assumed to be a decreasing function of

price. That is, price increases result in lower demand, and price decreases result inhigher demand (see Figure 2).

0 … p … 20.

x = f1p2 = 500120 - p2 0 … p … 20


p2010

x

10,000

Priceincreases

Pricedecreases

Demanddecreases

Demandincreases

x � f (p)� 500(20 � p)D

eman

d

Price

FIGURE 2

Economists use the elasticity of demand to study the relationship between changesin price and changes in demand. The elasticity of demand is the negative of the ratioof the relative rate of change of demand to the relative rate of change of price. Ifprice and demand are related by a price–demand equation of the form then the elasticity of demand can be expressed as

(3) = -

pf¿1p2f1p2

= -

f¿1p2f1p2

1p

-relative rate of change of demand

relative rate of change of price= -

d

dp ln f1p2d

dp ln p

x = f1p2,

Since p and f(p) are nonnegative and is negative (remember, demand is usually adecreasing function of price), (3) is always nonnegative. This is the reason that elasticityof demand was defined as the negative of a ratio.

f¿1p2I N S I G H T

Summarizing the preceding discussion, we have Theorem 1.

THEOREM 1 Elasticity of DemandIf price and demand are related by then the elasticity of demand is given by

E1p2 = -

pf¿1p2f1p2

x = f1p2,

The next example illustrates interpretations of the elasticity of demand.



E X A M P L E 2 Elasticity of Demand Find E(p) for the price–demand equation

Find and interpret each of the following:

(A) E(4) (B) E(16) (C) E(10)

SOLUTION

In order to interpret values of E(p), we must recall the definition of elasticity:

or

(A) If the $4 price changes by 10%, then the demand willchange by approximately

(B) If the $16 price changes by 10%, then the demand willchange by approximately

(C) If the $10 price changes by 10%, then the demand will alsochange by approximately 10%.E1102 =

1010 = 1.

4110%2 = 40%.E1162 =

164 = 4 7 1.

0.25110%2 = 2.5%.E142 =

416 = 0.25 6 1.

- a relative rate ofchange of demand

b L E1p2a relative rate ofchange of price

b

E1p2 = -

relative rate of change of demand

relative rate of change of price

E1p2 = -

pf¿1p2f1p2 = -

p1-5002500120 - p2 =

p

20 - p

x = f1p2 = 500120 - p2

Do not be concerned with the omission of any negative signs in these interpretations. Wealready know that if price increases, then demand decreases, and vice versa (Fig. 2). E(p)is a measure of how much the demand changes for a given change in price.

I N S I G H T

MATCHED PROBLEM 2 Find E(p) for the price–demand equation

Find and interpret each of the following:

(A) E(8) (B) E(30) (C) E(20)

x = f1p2 = 1,000140 - p2

The three cases illustrated in the solution to Example 2 are referred to as inelasticdemand, elastic demand, and unit elasticity, as indicated in Table 2.

E(p) Demand Interpretation

Inelastic Demand is not sensitive tochanges in price. A change inprice produces a smaller changein demand.

Elastic Demand is sensitive to changes inprice. A change in price producesa larger change in demand.

Unit A change in price produces thesame change in demand.

E1p2 = 1

E1p2 7 1

0 6 E1p2 6 1

TABLE 2

Now we want to see how revenue and elasticity are related. We have used the fol-lowing model for revenue many times before:

Revenue = 1demand2 * 1price2


Since we are looking for a connection between E(p) and revenue, we will use aprice–demand equation that has been written in the form where x is de-mand and p is price.

Revenue as a function of price

(4)

Since it follows from (4) that and always have thesame sign (see Table 3).

31 - E1p24R¿1p2x = f1p2 7 0,

= f1p231 - E1p24 = f1p2cpf¿1p2

f1p2 + 1 d E1p2 = -

pf ¿1p2f1p2

R¿1p2 = pf¿1p2 + f1p2 R1p2 = xp = pf1p2

x = f1p2,


All Are True or All Are False All Are True or All Are False

Demand is inelastic Demand is elastic

E(p) 7 1E(p) 6 1

R¿(p) 6 0R¿(p) 7 0

TABLE 3 Revenue and Elasticity of Demand

These facts are interpreted in the following summary and in Figure 3:

Revenue and Elasticity of Demand

Demand is inelastic:

A price increase will increase revenue.A price decrease will decrease revenue.

Demand is elastic:

A price increase will decrease revenue.A price decrease will increase revenue.

We know that a price increase will decrease demand at all price levels (Fig. 2). As Figure 3illustrates, the effects of a price increase on revenue depends on the price level. If de-mand is elastic, a price increase decreases revenue. If demand is inelastic, a price increaseincreases revenue.

I N S I G H T

p

R(p)

Priceincreases

Pricedecreases

Pricedecreases Revenue

decreasesRevenuedecreases

Revenueincreases

Priceincreases

Revenueincreases

Inelastic demand E(p) � 1Increasing revenue R�(p) � 0

Elastic demand E(p) � 1Decreasing revenue R�(p) � 0

FIGURE 3 Revenue and elasticity


Answers to Matched Problems 1. p1t2 =

100t + 100


E X A M P L E 3 Elasticity and Revenue A manufacturer of sunglasses currently sells one typefor $4 a pair. The price p and the demand x for these glasses are related by

If the current price is increased, will revenue increase or decrease?

SOLUTION

At the $4 price level, demand is inelastic and a price increase will increase revenue.

E142 =

410

= 0.4

=

p

14 - p

= -

p1-50027,000 - 500p

E1p2 = -

pf¿1p2f1p2

x = f1p2 = 7,000 - 500p

MATCHED PROBLEM 3 Repeat Example 3 if the current price for sunglasses is $10 a pair.

t125

p(t)

1

2.

(A) demand is inelastic.

(B) demand is elastic.

(C) demand has unit elasticity.

3. demand is elastic. Increasing price will decrease revenue.E1102 = 2.5;

E1202 = 1;

E1302 = 3;

E182 = 0.25;

E1p2 =

p

20 - p

Exercise 4-7

A In Problems 1–12, find the relative rate of change of f(x).

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12. f1x2 = 15x + 2x ln xf1x2 = 25x + 3x ln x

f1x2 = 5 - 3e-xf1x2 = 4 + 2e-2x

f1x2 = 50x - 0.01x2f1x2 = 100x - 0.5x2

f1x2 = 5x + 200f1x2 = 10x + 500

f1x2 = 4xf1x2 = 30x

f1x2 = 150f1x2 = 25

B In Problems 13–16, use the price–demand equation todetermine whether demand is elastic, is inelastic, or hasunit elasticity at the indicated values of p.

13.

(A) (B) (C)

14.

(A) (B) (C) p = 40p = 25p = 15

x = f1p2 = 1,875 - p2

p = 30p = 20p = 10

x = f1p2 = 12,000 - 10p2



15.

(A) (B) (C)

16.

(A) (B) (C)

17. Given the price–demand equation

(A) Express the demand x as a function of the price p.(B) Find the elasticity of demand, E(p).(C) What is the elasticity of demand when

If this price is increased by 10%, what is theapproximate change in demand?

(D) What is the elasticity of demand when If this price is increased by 10%, what is theapproximate change in demand?

(E) What is the elasticity of demand when If this price is increased by 10%, what is theapproximate change in demand?


(A) Express the demand x as a function of the price p.(B) Find the elasticity of demand, E(p).(C) What is the elasticity of demand when

If this price is decreased by 5%, what is theapproximate change in demand?

(D) What is the elasticity of demand when If this price is decreased by 5%, what is theapproximate change in demand?

(E) What is the elasticity of demand when If this price is decreased by 5%, what is theapproximate change in demand?


(A) Express the demand x as a function of the price p.(B) Express the revenue R as a function of the price p.(C) Find the elasticity of demand, E(p).(D) For which values of p is demand elastic? Inelastic?(E) For which values of p is revenue increasing?

Decreasing?(F) If and the price is decreased, will revenue

increase or decrease?(G) If and the price is decreased, will revenue

increase or decrease?p = $40

p = $10

0.02x + p = 60

p = $25?

p = $45?

p = $10?

p + 0.01x = 50

p = $15?

p = $25?

p = $10?

p + 0.005x = 30

p = 100p = 70p = 50

x = f1p2 = 875 - p - 0.05p2

p = 70p = 50p = 30

x = f1p2 = 950 - 2p - 0.1p2 20. Repeat Problem 19 for the price–demand equation

In Problems 21–26, use the price–demand equation to find thevalues of p for which demand is elastic and the valuesfor which demand is inelastic.

21.

22.

23.

24.

25.

26.

In Problems 27–32, use the demand equation to find therevenue function. Sketch the graph of the revenue function,and indicate the regions of inelastic and elastic demand on thegraph.

27.

28.

29.

30.

31.

32.

C If a price–demand equation is solved for p, then price isexpressed as and x becomes the independent variable.In this case, it can be shown that the elasticity of demand isgiven by

In Problems 33–36, use the price–demand equation to findE(x) at the indicated value of x.

33.

34.

35.

36.

37. Find E(p) for where A and k arepositive constants.

38. Find E(p) for where A and k arepositive constants.

x = f1p2 = Ae-kp,

x = f1p2 = Ap-k,

p = g1x2 = 20 - 2x, x = 100

p = g1x2 = 50 - 22x, x = 400

p = g1x2 = 30 - 0.05x, x = 400

p = g1x2 = 50 - 0.1x, x = 200

E1x2 = -

g1x2xg¿1x2

p = g1x2

x = f1p2 = 30 - 52p

x = f1p2 = 30 - 102p

x = f1p2 = 101p - 922x = f1p2 = 401p - 1522x = f1p2 = 10116 - p2x = f1p2 = 20110 - p2

x = f1p2 = 23,600 - 2p2

x = f1p2 = 22,500 - 2p2

x = f1p2 = 2324 - 2p

x = f1p2 = 2144 - 2p

x = f1p2 = 51p - 6022x = f1p2 = 101p - 3022

0.025x + p = 50

Applications39. Rate of change of cost. A fast-food restaurant can

produce a hamburger for $1.25. If the restaurant’s daily sales are increasing at the rate of 20 hamburgers per day, how fast is its daily cost for hamburgers increasing?

40. Rate of change of cost. The fast-food restaurant inProblem 39 can produce an order of fries for $0.40.If the restaurant’s daily sales are increasing at the

rate of 15 per day, how fast is its daily cost for fries increasing?

41. Revenue and elasticity. The price–demand equation forhamburgers at a fast-food restaurant is

Currently, the price of a hamburger is $2.00. If the price isincreased by 10%, will revenue increase or decrease?

x + 400p = 2,000


42. Revenue and elasticity. Refer to Problem 41. If the currentprice of a hamburger is $3.00, will a 10% price increasecause revenue to increase or decrease?

43. Revenue and elasticity. The price–demand equation foran order of fries at a fast-food restaurant is

Currently, the price of an order of fries is $0.30. If theprice is decreased by 10%, will revenue increase ordecrease?

44. Revenue and elasticity. Refer to Problem 43. If the currentprice of an order of fries is $0.60, will a 10% price decreasecause revenue to increase or decrease?

45. Maximum revenue. Refer to Problem 41. What price willmaximize the revenue from selling hamburgers?

46. Maximum revenue. Refer to Problem 43. What price willmaximize the revenue from selling fries?

47. Population growth. A model for Canada’s populationgrowth (Table 4) is

where t is years since 1950. Find and graph the percentagerate of change of f(t) for 0 … t … 50.

f1t2 = 0.34t + 14.6

x + 1,000p = 1,000

48. Population growth. A model for Mexico’s populationgrowth (Table 4) is

where t is years since 1950. Find and graph the percentagerate of change of f(t) for

49. Crime. A model for the number of robberies in theUnited States (Table 5) is

where t is years since 1990. Find the relative rate ofchange for robberies in 2002.

r1t2 = 11.3 - 3.6 ln t

0 … t … 50.

f1t2 = 1.47t + 25.5


Canada Mexico Year (millions) (millions)

1950 14 28

1960 18 39

1970 22 53

1980 25 68

1990 28 85

2000 31 100

TABLE 4 Population Growth

Robbery Aggravated Assault

1995 5.4 9.5

1996 5.2 8.8

1997 4.3 8.6

1998 4.0 7.5

1999 3.6 6.7

2000 3.2 5.7

2001 2.8 5.3

2002 2.2 4.3

TABLE 5 Number of Victimizations per 1,000Population Age 12 and Over

50. Crime. A model for the number of assaults in the UnitedStates (Table 5) is

where t is years since 1990. Find the relative rate ofchange for assaults in 2002.

a1t2 = 19.6 - 6.0 ln t

CHAPTER 4 REVIEW

4-1 The Constant e and Continuous Compound Interest• The number e is defined as

If the number of compounding periods in one year is increased without limit, we obtain the compoundinterest formula

where

A = amount at time t t = time in years r = annual nominal interest rate compounded continuously

P = principal

A = Pert

limx: q

a1 +

1nbn

= limx:011 + s21>s = 2.718 281 828 459 Á

Ex. 1, p. 219Ex. 2, p. 219Ex. 3, p. 220Ex. 4, p. 221

Examples

Important Terms, Symbols, and Concepts


4-2 Derivatives of Exponential and Logarithmic Functions• For

• The change-of-base formulas allow conversion from base e to any base

4-3 Derivatives of Products and Quotients• Product rule. If

• Quotient rule. If then provided that both

and

4-4 The Chain Rule• A function m is a composite of functions f and g if

• The chain rule gives a formula for the derivative of the composite function

• A special case of the chain rule is the general power rule:

• Other special cases of the chain rule are the following general derivative rules:

4-5 Implicit Differentiation• If is a function defined by the equation we use implicit differentiation to find an

equation in x, y, and

4-6 Related Rates• If x and y represent quantities that are changing with respect to time and are related by the equation

then implicit differentiation produces an equation that relates x, y, and Problems of this type are called related-rates problems.

• Suggestions for solving related-rates problems are given on page 258.

4-7 Elasticity of Demand• The relative rate of change, or the logarithmic derivative, of a function f(x) is , and the

percentage rate of change is .

• If price and demand are related by then the elasticity of demand is given by

• Demand is inelastic if (Demand is not sensitive to changes in price; a change in priceproduces a smaller change in demand.) Demand is elastic if (Demand is sensitive to changesin price; a change in price produces a larger change in demand.) Demand has unit elasticity if (A change in price produces the same change in demand.)

• If is the revenue function, then and always have the same sign.31 - E1p24R¿1p2R1p2 = pf1p2E1p2 = 1.

E1p2 7 1.0 6 E1p2 6 1.

E1p2 = -

pf¿1p2f1p2 = -

relative rate of change of demand

relative rate of change of price

x = f1p2,100 * [f¿1x2>f1x2]

f¿1x2>f1x2

dx>dt.dy>dt,F1x, y2 = 0,

y¿.F1x, y2 = 0,y = y1x2

d

dx ef1x2

= ef1x2 f¿1x2

d

dx ln [f1x2] =

1f1x2 f¿1x2

d

dx [f1x2]n

= n[f1x2]n - 1f¿1x2

m¿1x2 = f¿[g1x2]g¿1x2m1x2 = f[g1x2]:

m1x2 = f[g1x2].

B¿1x2 exist.

T¿1x2f¿1x2 =

B1x2 T¿1x2 - T1x2 B¿1x2CB1x2 D 2y = f1x2 =

T1x2B1x2,

F¿1x2 and S¿1x2 exist.provided that bothy = f1x2 = F1x2 S1x2, then f¿1x2 = F1x2S¿1x2 + S1x2F¿1x2,

bx= ex ln b logb x =

ln xln b

b, b 7 0, b Z 1:

d

dx ln x =

1x

d

dx logb x =

1ln b

1x

d

dx ex

= ex d

dx bx

= bx ln b

b 7 0, b Z 1,

Chapter 4 Review 271

Ex. 1, p. 225Ex. 2, p. 227Ex. 3, p. 228Ex. 4, p. 229Ex. 5, p. 230

Ex. 1, p. 233Ex. 2, p. 234Ex. 3, p. 234Ex. 4, p. 236Ex. 5, p. 237Ex. 6, p. 237

Ex. 1, p. 241Ex. 2, p. 241Ex. 4, p. 245Ex. 5, p. 247Ex. 3, p. 243

Ex. 6, p. 247

Ex. 1, p. 253Ex. 2, p. 254Ex. 3, p. 255

Ex. 1, p. 257Ex. 2, p. 258Ex. 3, p. 259

Ex. 1, p. 263

Ex. 2, p. 265

Ex. 3, p. 268



Work through all the problems in this chapter review, andcheck your answers in the back of the book. Answers to allreview problems are there, along with section numbers in ital-ics to indicate where each type of problem is discussed. Whereweaknesses show up, review appropriate sections of the text.

A1. Use a calculator to evaluate to the nearest

cent for and 20.

Find the indicated derivatives in Problems 2–4.

2. 3.

4. for

5. Let and

(A) Express y in terms of x.

(B) Use the chain rule to find dy�dx, and then expressdy�dx in terms of x.

6. Find for defined implicity by the equationand evaluate at

7. For where and finddy�dt if when

8. Given the demand equation

(A) Express the demand x as a function of the price p.

(B) Find the elasticity of demand, E(p).

(C) Find E(15) and interpret.

(D) Express the revenue function as a function of price p.

(E) If what is the effect of a price cut on revenue?

B9. Find the slope of the line tangent to when


Do you think the limit exists? If so, what do you think it is?


11. 12.

13. 14. for

15. for

16. dy�dx for

17. Find the equation of the line tangent to the graphof at At

18. Find for defined implicitly by the equationand find the slope of the graph at

1-1, 22.x2

- 3xy + 4y2= 23,

y = y1x2y¿

x = -1.x = 0.y = f1x2 = 1 + e-x

y = e-2x ln 5x

f1x2 = ex3- x2

f¿1x2y = ln12x3

- 3x2y¿

d

dx ex

x6

d

dx1x6 ln x2d

dz 31ln z27 + ln z74

limn: q

a1 +

2nbn

x = 0.y = 100e-0.1x

p = $25,

25p + x = 1,000,

x = 12.dx>dt = 3y = y1t2,x = x1t2y = 3x2

- 5,

1x, y2 = 11, 22.2y2- 3x3

- 5 = 0,y = y1x2y¿

u = 3 + ex.y = ln u

y = ln12x + 72y¿

d

dx e2x - 3d

dx12 ln x + 3ex2

t = 5, 10,A = 2,000e0.09t

19. Find for defined implicitly byand evaluate at

20. Find for defined implicitly by and evaluate at (1, 0).

21. Find for defined implicitly by and evaluate at (1, 1).

22. A point is moving on the graph of so that itsx coordinate is decreasing by 2 units per second when

Find the rate of change of the y coordinate.

23. A 17-foot ladder is placed against a wall. If the foot of theladder is pushed toward the wall at 0.5 foot per second,how fast is the top of the ladder rising when the foot is8 feet from the wall?

24. Water from a water heater is leaking onto a floor. A circu-lar pool is created whose area is increasing at the rate of24 square inches per minute. How fast is the radius R ofthe pool increasing when the radius is 12 inches?

C25. Find the values of p for which demand is elastic and the

values for which demand is inelastic if the price–demandequation is

26. Graph the revenue function and indicate the regionsof inelastic and elastic demand of the graph if theprice–demand equation is

27. Let and

(A) Express y in terms of x.

(B) Use the chain rule to find dy�dx, and then expressdy�dx in terms of x.


28. for

29.

30.

31. Find for defined implicitly by the equationand evaluate at (0, 0).

32. A rock thrown into a still pond causes a circular ripple.Suppose the radius is increasing at a constant rate of 3feet per second. Show that the area does not increase at aconstant rate. When is the rate of increase of the area thesmallest? The largest? Explain.

33. A point moves along the graph of in such a waythat its y coordinate is increasing at a constant rate of 5units per second. Does the x coordinate ever increase at afaster rate than the y coordinate? Explain.

y = x3

exy= x2

+ y + 1,y = y1x2y¿

d

dx2ln1x2

+ x2

d

dx log51x2

- x2y = 5x2

- 1y¿

u = 4 - ex.y = w3, w = ln u,

x = f1p2 = 5120 - p2 0 … p … 20

x = f1p2 = 201p - 1522 0 … p … 15

3A = pR24

1x, y2 = 11, 42.y2

- 4x2= 12

ln y = x2- y2,y = y1x2y¿

x - y2= ey,y = y1x2y¿

1t, x2 = 1-2, 22.x3- 2t2x + 8 = 0,

x = x1t2x¿

REVIEW EXERCISE


Applications 273

34. Doubling time. How long will it take money to double if it is invested at 5% interest compounded

(A) Annually? (B) Continuously?

35. Continuous compound interest. If $100 is invested at 10% interest compounded continuously, the amount (in dollars) at the end of t years is given by

Find and

36. Marginal analysis. The price–demand equation for 14-cubic-foot refrigerators at an appliance store is

where x is the monthly demand and p is the price indollars. Find the marginal revenue equation.

37. Demand equation. Given the demand equation

find the rate of change of p with respect to x by implicitdifferentiation (x is the number of items that can be sold at a price of $p per item).

38. Rate of change of revenue. A company is manufacturing anew video game and can sell all that it manufactures. Therevenue (in dollars) is given by

where the production output in 1 day is x games. If pro-duction is increasing at 10 games per day when produc-tion is 250 games per day, find the rate of increase inrevenue.

39. Revenue and elasticity. The price–demand equation forhome-delivered 12-inch pizzas is

where x is the number of pizzas delivered weekly. Thecurrent price of one pizza is $8. In order to generate addi-tional revenue from the sale of 12-inch pizzas, would yourecommend a price increase or a price decrease?

p = 16.8 - 0.002x

R = 36x -

x2

20

x = 25,000 - 2p3

p1x2 = 1,000e-0.02x

A¿1102.A¿1t2, A¿112,A = 100e0.1t

40. Average income. A model for the average income perhousehold before taxes are paid is

where t is years since 1980. Find the relative rate ofchange of household income in 2010.

41. Drug concentration. The concentration of a drug in the bloodstream t hours after injection is given approxi-mately by

where C(t) is concentration in milligrams per milliliter.What is the rate of change of concentration after 1 hour?After 5 hours?

42. Wound healing. A circular wound on an arm is healing atthe rate of 45 square millimeters per day (the area of thewound is decreasing at this rate). How fast is the radius Rof the wound decreasing when millimeters?

43. Psychology: learning. In a computer assembly plant, anew employee, on the average, is able to assemble

units after t days of on-the-job training.

(A) What is the rate of learning after 1 day? After5 days?

(B) Find the number of days (to the nearest day) afterwhich the rate of learning is less than 0.25 unit per day.

44. Learning. A new worker on the production line performsan operation in T minutes after x performances of theoperation, as given by

If, after performing the operation 9 times, the rate of improvement is operations per hour, find therate of improvement in time dT�dt in performing eachoperation.

dx>dt = 3

T = 2a1 +

1

x3>2 b

N1t2 = 1011 - e-0.4t2

3A = pR24R = 15

C1t2 = 5e-0.3t

f1t2 = 1,700t + 20,500

APPLICATIONS


S e c t i o n 6 - 1 Basic Concepts 275


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